The terms "sr" and "lr" cost functions typically refer to "short-run" and "long-run" cost functions in economics. The distinction between the two depends on the time horizon over which the costs are being considered.
In the short run, some inputs are fixed and cannot be changed, while others are variable and can be adjusted. For example, in the short run, a factory may have fixed costs such as rent, property taxes, and insurance, while variable costs may include labor, raw materials, and utilities. The short-run cost function reflects how the total cost of production changes as the variable inputs are increased or decreased while the fixed inputs remain constant.
In the long run, all inputs are variable and can be adjusted. For example, in the long run, a factory may be able to build a larger building, buy more machines, or relocate to a cheaper area. The long-run cost function reflects how the total cost of production changes as all inputs are increased or decreased.
An example of a short-run cost function could be the cost of producing bread in a bakery, where the cost of flour, yeast, and electricity are variable costs, but the cost of rent for the bakery building is a fixed cost.
An example of a long-run cost function could be the cost of running a transportation company, where the cost of vehicles, fuel, and labor are all variable costs, but the cost of building a new headquarters or expanding the business into a new market are fixed costs.
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High power microwave tubes used for satellite communications have lifetimes that follow an exponential distribution with E[X] =3 years: (a) (3 points) What is the probability that the life of a tube will exceed 4 years ?
The probability that the life of a tube will exceed 4 years is approximately 0.2636 or 26.36%.
Since the lifetime of a tube follows an exponential distribution with a mean of 3 years, we can use the exponential distribution formula:
f(x) = λe^(-λx)
where λ is the rate parameter, which is the inverse of the mean, λ = 1/3.
To find the probability that the life of a tube will exceed 4 years, we need to integrate the PDF from x = 4 to infinity:
P(X > 4) = ∫_4^∞ λe^(-λx) dx
= [-e^(-λx)]_4^∞
= e^(-4λ)
= e^(-4/3)
≈ 0.2636
Therefore, the probability that the life of a tube will exceed 4 years is approximately 0.2636 or 26.36%.
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let z = a bi and w = c di. prove the following property: ez ew = ez w . 6
We have proved the property ez ew = ez+w.
To prove the property ez ew = ez+w, where z = a + bi and w = c + di, we can use the properties of complex exponentials.
First, let's express ez and ew in their exponential form:
ez = e^(a+bi) = e^a * e^(ib)
ew = e^(c+di) = e^c * e^(id)
Now, we can multiply ez and ew together:
ez ew = (e^a * e^(ib)) * (e^c * e^(id))
Using the properties of exponentials, we can simplify this expression:
ez ew = e^a * e^c * e^(ib) * e^(id)
Now, we can use Euler's formula, which states that e^(ix) = cos(x) + i sin(x), to express the complex exponentials in terms of trigonometric functions:
e^(ib) = cos(b) + i sin(b)
e^(id) = cos(d) + i sin(d)
Substituting these values back into the expression, we get:
ez ew = e^a * e^c * (cos(b) + i sin(b)) * (cos(d) + i sin(d))
Using the properties of complex numbers, we can expand and simplify this expression:
ez ew = e^a * e^c * (cos(b)cos(d) - sin(b)sin(d) + i(sin(b)cos(d) + cos(b)sin(d)))
Now, let's express ez+w in exponential form:
ez+w = e^(a+bi+ci+di) = e^((a+c) + (b+d)i)
Using Euler's formula again, we can express this exponential in terms of trigonometric functions:
ez+w = e^(a+c) * (cos(b+d) + i sin(b+d))
Comparing this with our previous expression for ez ew, we can see that they are equal:
ez ew = e^a * e^c * (cos(b)cos(d) - sin(b)sin(d) + i(sin(b)cos(d) + cos(b)sin(d))) = e^(a+c) * (cos(b+d) + i sin(b+d)) = ez+w
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Given begin mathsize 18px style sin theta equals 2 over 5 end style, find begin mathsize 18px style cos theta end style if it is in the first quadrant. 0. 6
0. 84
0. 4
0. 92
The cos(θ) is approximately 0.92.
To solve this problemWe can use the Pythagorean identity to find cos(θ).
The Pythagorean identity states that [tex]sin^2[/tex](θ) + [tex]cos^2[/tex] (θ) = 1.
Given sin(θ) = 2/5, we can substitute this value into the equation:
[tex](2/5)^2 + cos^2(\pi) = 14/25 + cos^2(\pi) = 1[/tex]
Now, we can solve for
[tex]cos^2 (\theta): cos^2[\theta] = 1 - 4/25cos^2(\theta) = 25/25 - 4/25[tex]cos^2(\theta) = 21/25[/tex]
Taking the square root of both sides, we get:
cos(θ) = ± [tex]\sqrt(21/25)[/tex]
Since θ is in the first quadrant, we take the positive value:cos(θ) = sqrt(21/25)
Simplifying further:
cos(θ) = [tex]\sqrt(21)/\sqrt(25)[/tex]cos(θ) = sqrt(21)/5
Approximating the value of [tex]\sqrt(21)[/tex] to two decimal places:cos(θ) ≈ 0.92
Therefore, cos(θ) is approximately 0.92.
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Consider the hypothesis testH_0:\mu_1=\mu_2againstH_1:\mu_1\neq \mu_2with known variances\sigma _1=9and\sigma _2=6. Suppose that sample sizesn_1=11andn_2=14and that\overline{x}_1=4.7and\overline{x}_2=7.8Useg= 0,05(a) Test the hypothesis and find the P-value.(b) What is the power of the test in part (a) for a true difference in means of 3?(c) Assuming equal sample sizes, what sample size should be used to obtain\beta =0.05if the true difference in means is 3? Assume that(a) The null hypothesis Choose your answer; The null hypothesis _ rejected rejected. The P-value is Enter your answer; The P-value is . Round your answer to three decimal places (e.g. 98.765).
(b) The power is Enter your answer in accordance to the item b) of the question statement
. Round your answer to two decimal places (e.g. 98.76).
(c) Enter your answer in accordance to the item c) of the question statement . Round your answer up to the nearest integer.
a) We reject the null hypothesis at the 0.05 level of significance.
b) The power of the test is approximately 0.666 or 66.6%.
c) A sample size of at least 47 to achieve a power of 0.95 when the true difference in means is 3.
(a) To test the hypothesis, we can use a two-sample t-test. The test statistic is given by:
[tex]t = (\overline{x}_1 - \overline{x}_2) / \sqrt{ ( \sigma_1^2/n_1 ) + ( \sigma_2^2/n_2 ) }[/tex]
Plugging in the values given, we get:
[tex]t = (4.7 - 7.8) / \sqrt{ ( 9/11 ) + ( 6/14 ) } = -3.206[/tex]
The degrees of freedom for this test are df = n1 + n2 - 2 = 23. Using a t-table or calculator, we find that the P-value is less than 0.005.
(b) The power of the test is the probability of rejecting the null hypothesis when the alternative hypothesis is true. In this case, the alternative hypothesis is that the true difference in means is 3. We can use the non-central t-distribution to calculate the power:
[tex]power = 1 - P( |t| < t_{1-\alpha/2,n_1+n_2-2,\delta} )[/tex]
where[tex]t_{1-\alpha/2,n_1+n_2-2,\delta}[/tex]is the critical value of the t-distribution with n1 + n2 - 2 degrees of freedom, a significance level of 0.05, and a non-centrality parameter of
Plugging in the values given, we get:
δ =[tex](3) / \sqrt{ ( 9/11 ) + ( 6/14 ) } = 2.198[/tex]
[tex]t_{1-\alpha/2,n_1+n_2-2,\delta} = +2.074[/tex]
Therefore, the power of the test is:
power = 1 - P( |t| < 2.074 )
Using a t-table or calculator with 23 degrees of freedom and a non-centrality parameter of 2.198, we find that P( |t| < 2.074 ) ≈ 0.334.
(c) Assuming equal sample sizes, we can use the following formula to find the sample size needed to achieve a power of 0.95:
[tex]n = [ (z_{1-\beta/2} + z_{1-\alpha/2}) / δ ]^2[/tex]
where[tex]z_{1-\beta/2} and z_{1-\alpha/2}[/tex] are the critical values of the standard normal distribution for a power of 0.95 and a significance level of 0.05, respectively.
Plugging in the values given, we get:
δ =[tex](3) / \sqrt{ ( 9/n ) + ( 6/n ) } = 0.925[/tex]
[tex]z_{1-\beta/2} = 1.96[/tex] (for a power of 0.95)
[tex]z_{1-\alpha/2} = 1.96[/tex]
Solving for n, we get:
[tex]n = [ (1.96 + 1.96) / 0.925 ]^2 = 46.24[/tex]
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Bev had 24 pieces of candy she gave Jimmy 1/3 from the candy pieces remaining then she gave Selena 1/4 how many pieces of candy does she have left
After giving Jimmy one-third of the remaining candy pieces and Selena one-fourth of the remaining candy pieces, Bev is now down to having two-thirds as many as three-quarters as many as twenty-four pieces of candy.
Calculating how much candy is still available after each distribution is necessary if we want to establish how many pieces of candy Bev still possesses. At the beginning, Bev has twenty-four individual bits of candy. After giving Jimmy a third of the candy pieces, the number of pieces that are still remaining may be computed as (2/3) times 24, which is equal to two-thirds of the total amount.
The next thing that happens is that Bev gives Selena a quarter of the remaining candy pieces. We need to multiply the total amount that is still available by one quarter since Selena is entitled to a portion of what is left over after Jimmy has received his part. As a result, the remaining candy pieces can be approximated using the formula (3/4 * (2/3) * 24 after Selena has been given her portion.
The solution to the equation is found to be (3/4) * (2/3) * 24, which is 4 * 8, which equals 32. Therefore, after giving Jimmy one third of the remaining candy pieces and Selena one quarter of the remaining candy pieces, Bev still has 32 pieces of candy left.
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(Q14 in book): Larry Ellison starts a company that manufacturers high-end custom leather bags. He hires two employees. Each employee only begins working on a bag when a customer order has been received and then she makes the bag from beginning to end. The average production time of a bag is 1. 8 days with a standard deviation of 2. 7 days. Larry expects to receive one customer order per day on average. The inter-arrival times of orders have a coefficient of variation of 1. The expected duration, in days, between when an order is received and when production begins on the bag, equals: ______________________ [days]. (Note, this duration includes the time waiting to start production but do not include the time in production. ) Question 5 options:
The expected duration, in days, between when an order is received and when production begins on the bag is 2.25 days.
Larry Ellison has started a company that manufactures high-end custom leather bags and he has hired two employees. Each employee only starts working on a bag when a customer order has been received and then she makes the bag from beginning to end.
The average production time of a bag is 1.8 days with a standard deviation of 2.7 days. Larry expects to receive one customer order per day on average.
The inter-arrival times of orders have a coefficient of variation of 1.
To calculate the expected duration, use the following formula: Expected duration = (1/λ) - (1/μ)
where λ is the arrival rate and μ is the average processing time per item.
Substituting the given values, we have:λ = 1 per dayμ = 1.8 days Expected duration = (1/1) - (1/1.8)
Expected duration = 0.56 days or 2.25 days (rounded to two decimal places)Therefore, the expected duration, in days, between when an order is received and when production begins on the bag is 2.25 days.
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Find the Taylor series, centered at c= 7, for the function 1 f(x) = 2 Q f(x) = n=0 The interval of convergence is:
Find the Taylor series, centered at c=7c=7, for the function
f(x)=1x.f(x)=1x.
f(x)=∑n=0[infinity]f(x)=∑n=0[infinity] .
The interval of convergence is:
The Taylor series expansion for the function f(x) = 1/x centered at c = 7 is given by the infinite sum:
f(x) = 1/7 - 1/49(x-7) + 1/343(x-7)² - 1/2401(x-7)³ + ...
And the interval of convergence for this series is (7 - R, 7 + R),
To find the Taylor series for a function, we start by calculating the derivatives of the function at the center point (c) and evaluating them at c. In this case, we have f(x) = 1/x, so let's begin by finding the derivatives:
f(x) = 1/x f'(x) = -1/x² (derivative of 1/x)
f''(x) = 2/x³ (derivative of -1/x²)
f'''(x) = -6/x^4 (derivative of 2/x³)
f''''(x) = 24/x⁵ (derivative of -6/x⁴) ...
We can observe a pattern in the derivatives of f(x). The nth derivative of f(x) can be written as (-1)ⁿ⁺¹ * n! / xⁿ⁺¹, where n! represents the factorial of n.
Now, we can use these derivatives to construct the Taylor series expansion. The general form of the Taylor series for a function f(x) centered at c is given by:
f(x) = f(c) + f'(c)(x-c) + f''(c)(x-c)²/2! + f'''(c)(x-c)³/3! + ...
In our case, the center point is c = 7. Let's substitute the values into the series:
f(x) = f(7) + f'(7)(x-7) + f''(7)(x-7)²/2! + f'''(7)(x-7)³/3! + ...
To find the coefficients, we need to evaluate the derivatives at c = 7:
f(7) = 1/7 f'(7) = -1/49 f''(7) = 2/343 f'''(7) = -6/2401 ...
Plugging these values into the series, we get:
f(x) = 1/7 - 1/49(x-7) + 2/343(x-7)²/2! - 6/2401(x-7)³/3! + ...
Simplifying further:
f(x) = 1/7 - 1/49(x-7) + 1/343(x-7)² - 1/2401(x-7)³ + ...
Now, let's talk about the interval of convergence for this Taylor series. The interval of convergence refers to the range of values of x for which the Taylor series accurately represents the original function. In this case, the function f(x) = 1/x is not defined at x = 0.
Therefore, the interval of convergence for this Taylor series is (7 - R, 7 + R), where R is the distance from the center point to the nearest singularity (in this case, x = 0).
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The temperature in town is "-12. " eight hours later, the temperature is 25. What is the total change during the 8 hours?
The temperature change is the difference between the final temperature and the initial temperature. In this case, the initial temperature is -12, and the final temperature is 25. To find the temperature change, we simply subtract the initial temperature from the final temperature:
25 - (-12) = 37
Therefore, the total change in temperature over the 8-hour period is 37 degrees. It is important to note that we do not know how the temperature changed over the 8-hour period. It could have gradually increased, or it could have changed suddenly. Additionally, we do not know the units of temperature, so it is possible that the temperature is measured in Celsius or Fahrenheit. Nonetheless, the temperature change remains the same, regardless of the units used.
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sketch the finite region enclosed by the given curves and find the area of the region. y=squarootx, y=x^2, x=2
The area of the region enclosed by the curves y = √x, y = x² and x = 2 is 4√2/4 - 8/3
To sketch the finite region enclosed by the curves y = √x, y = x² and x = 2 we can first plot the two functions and the vertical line
The region we are interested in is the shaded area between the two curves and to the left of the line x=2. To find the area of this region, we can integrate the difference between the two functions with respect to x over the interval [0] [2]
[tex]\int_0^2(\sqrt{x} -x^2)dx[/tex]
Evaluating this integral, we get:
= [tex][\frac{2}{3} x^{\frac{3}{2}}-\frac{1}{3} x^3]_0^2[/tex]
= [tex]\frac{2}{3} (2)^\frac{3}{2} - \frac{1}{3}(2)^3-0[/tex]
= 4√2/4 - 8/3
Therefore, the area of the region enclosed by the curves y = √x, y = x² and x = 2 is 4√2/4 - 8/3
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The index of refraction of crown glass is 1.53 for violet light, and it is 1.51 for red light. a. What is the speed of violet light in crown glass? b. What is the speed …The index of refraction of crown glass is 1.53 for violet light, and it is 1.51 for red light.a. What is the speed of violet light in crown glass?b. What is the speed of red light in crown glass?
The speed of red light in crown glass is approximately 1.99 x [tex]10^8[/tex] m/s.
a. The speed of violet light in crown glass can be calculated using the formula:
v = c/n
Where,
v is the speed of light in the material,
c is the speed of light in vacuum and
n is the index of refraction of the material for violet light.
Plugging in the values given, we get:
violet light speed in crown glass = c/n
violet light speed in crown glass = c/1.53
Using the value for the speed of light in vacuum,
c = 3.00 x [tex]10^8[/tex] m/s,
We can calculate the speed of violet light in crown glass as:
violet light speed in crown glass = (3.00 x [tex]10^8[/tex] m/s) / 1.53
= 1.96 x [tex]10^8[/tex] m/s
Therefore, the speed of violet light in crown glass is approximately 1.96 x [tex]10^8[/tex] m/s.
b. Similarly, the speed of red light in crown glass can be calculated using the same formula, but with the index of refraction for red light:
red light speed in crown glass = c/n = c/1.51
Using the same value for the speed of light in vacuum and plugging in the value for the index of refraction for red light, we get:
The red light speed in crown glass = (3.00 x [tex]10^8[/tex] m/s) / 1.51 = 1.99 x 10^8 m/s
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And are tangent to circle A. Kite ABCD is inscribed in circle E. The radius of circle A is 14 and the radius of circle E is 15. Find the length of rounded to the nearest tenth
The length of AD (rounded to the nearest tenth) cannot be found as there is some mistake in the given question or data.
Given that Kite ABCD is inscribed in circle E and are tangent to circle A. The radius of circle A is 14 and the radius of circle E is 15.
To find: The length of AD (rounded to the nearest tenth)Solution:Since ABCD is a kite, we can say that the two diagonals of the kite are perpendicular to each other.AD is one of the diagonals of the kite.
We need to find the length of AD to find its area, and then we will equate the area of kite ABCD to the product of its diagonals as a property of kite.
The other diagonal of the kite BD is a chord of circle E.The radius of circle E is 15 cmSo, the length of BD is 30 cm. (as it is the diameter of the circle E)Let's consider a right triangle AOD as shown below:
In triangle AOD,By Pythagoras theorem, we have:OD² + AD² = AO²
(where AO = radius of circle A = 14)
OD² + AD² = 14²
AD² = 14² - OD²
AD² = 196 - (15)²
AD² = 196 - 225
AD² = -29
AD = √(-29) (which is not possible as AD is a length and length cannot be negative)So, there is a mistake in the given question or data
Therefore, the given problem cannot be solved.
The length of AD (rounded to the nearest tenth) cannot be found as there is some mistake in the given question or data.
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describe geometrically the effect of the transformation t. let a = [0 0 0 0 1 0 0 0 1']
The transformation t applied to vector a rotates it by 90 degrees around the y-axis and then scales it by a factor of 2 along the x-axis.
The given vector a can be represented in 3D space as (0,0,0,0,1,0,0,0,1)^T, where T denotes the transpose.
To apply the rotation, we first represent the rotation matrix R about the y-axis by an angle of 90 degrees as:
R = [0 0 1 0 1 0 -1 0 0;
0 1 0 0 0 0 0 0 1;
-1 0 0 1 0 0 0 0 0]
Multiplying R with a, we get:
Ra = [0 0 1 0 1 0 -1 0 0]^T
This means that a is rotated by 90 degrees around the y-axis.
Next, we apply the scaling along the x-axis. We represent the scaling matrix S as:
S = [2 0 0;
0 1 0;
0 0 1]
Multiplying S with Ra, we get:
SRa = [0 0 2 0 1 0 -2 0 0]^T
This means that Ra is scaled by a factor of 2 along the x-axis.
Thus, the transformation t applied to vector a rotates it by 90 degrees around the y-axis and then scales it by a factor of 2 along the x-axis. Geometrically, this can be visualized as taking the original vector a and rotating it clockwise by 90 degrees about the y-axis, and then stretching it horizontally along the x-axis.
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TRUE/FALSE. The R command "qchisq(0.05,12)" is for finding the chi-square critical value with 12 degrees of freedom at alpha = 0.05.
In this case, the R command "qchisq(0.05,12)" returns the critical value of the chi-square distribution with 12 degrees of freedom at the probability level of 0.05, which is used to determine whether the test statistic falls in the rejection region or not in a statistical test.
True. The R command "qchisq(p, df)" is used to find the critical value of the chi-square distribution with "df" degrees of freedom at the specified probability level "p". In this case, "qchisq(0.05,12)" returns the critical value of the chi-square distribution with 12 degrees of freedom at the probability level of 0.05.
The chi-square distribution is a family of probability distributions that arise in many statistical tests, such as the chi-square test of independence, goodness of fit tests, and tests of association in contingency tables.
The distribution is defined by its degrees of freedom (df), which determines its shape and location. The critical value of the chi-square distribution is the value at which the probability of obtaining a more extreme value is equal to the specified level of significance (alpha).
Therefore, in this case, the R command "qchisq(0.05,12)" returns the critical value of the chi-square distribution with 12 degrees of freedom at the probability level of 0.05, which is used to determine whether the test statistic falls in the rejection region or not in a statistical test.
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A geometry set cost 29.75 together. david paid 228.55 for 12 geometry set and 5 calculators. how much would tim pay for 2 geometry sets and 4 calculators?
Tim would pay $144.5183 for 2 geometry sets and 4 calculators.
Given that the cost of a geometry set is 29.75 together.
David paid 228.55 for 12 geometry sets and 5 calculators.
To find out how much Tim would pay for 2 geometry sets and 4 calculators, we need to calculate the cost of the 2 geometry sets and 4 calculators.
So, 2 geometry sets cost = 2 × 29.75
= $59.505
calculators cost = 5/12 × 228.55 - 59.75
= $85.0183
Total cost of 2 geometry sets and 4 calculators is
= 59.5 + 85.0183= $144.5183
Therefore, Tim would pay $144.5183 for 2 geometry sets and 4 calculators.
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Consider the following. f(x, y, z) = Squareroot x + yz, P(1, 3, 1), u = (3/7, 6/7, 2/7) Find the gradient of f. nabla f(x, y, z) = Evaluate the gradient at the point P. nabla f(1, 3, 1) = Find the rate of change of f at P in the direction of the vector u. D_u f(1, 3, 1) =
The gradient of f is nabla f(x, y, z) = (1/sqrt(x+yz), z/sqrt(x+yz), y/sqrt(x+yz)).
At point P, the gradient is nabla f(1, 3, 1) = (1/2, 1/sqrt(2), sqrt(2)/2).
The rate of change of f at P in the direction of the vector u is D_u f(1, 3, 1) = 9/7sqrt(2).
The gradient of f is defined as the vector of partial derivatives of f with respect to its variables. Hence, we have nabla f(x, y, z) = (∂f/∂x, ∂f/∂y, ∂f/∂z) = (1/sqrt(x+yz), z/sqrt(x+yz), y/sqrt(x+yz)).
Substituting the values of P into this expression, we get nabla f(1, 3, 1) = (1/2, 1/sqrt(2), sqrt(2)/2).
The directional derivative of f at P in the direction of the unit vector u is given by the dot product of the gradient of f at P and the unit vector u, i.e., D_u f(1, 3, 1) = nabla f(1, 3, 1) · u.
Substituting the values of P and u into this expression, we get D_u f(1, 3, 1) = (1/2) * (3/7) + (1/sqrt(2)) * (6/7) + (sqrt(2)/2) * (2/7) = 9/7sqrt(2). Therefore, the rate of change of f at P in the direction of the vector u is 9/7sqrt(2).
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let x1 ... xn be a random sample from a n(μ 1) population. find the mle of μ. a) 1/x. b) 1/x^2. c) X. d) X-1.
The maximum likelihood estimator (MLE) of μ for a normal population with known variance is X, the sample mean. Therefore, the MLE of μ in this case is option (c), X.
The MLE is the value of the parameter that maximizes the likelihood function, which is the joint probability density function of the sample. For a random sample from a normal population with known variance, the likelihood function is proportional to exp(-1/2∑(xi-μ)^2/sigma^2), where the sum is taken over all the sample values xi. Taking the derivative of this function with respect to μ and setting it equal to zero, we obtain the equation X = μ, which implies that X is the MLE of μ. Option (a) and (b) do not make sense as they involve taking the inverse or inverse square of the sample, and option (d) suggests subtracting 1 from the sample mean, which is not a valid estimator for μ.
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find the general solution of the given system. dx dt = −9x 4y dy dt = − 5 2 x 2y
The general solution of the system is x(t) = Ce^(-9t), y(t) = De^(5C^2/36 e^(-18t)).
We have the system of differential equations:
x/dt = -9x
dy/dt = -(5/2)x^2 y
The first equation has the solution:
x(t) = Ce^(-9t)
where C is a constant of integration.
We can use this solution to find the solution for y. Substituting x(t) into the second equation, we get:
dy/dt = -(5/2)C^2 e^(-18t) y
Separating the variables and integrating:
∫(1/y) dy = - (5/2)C^2 ∫e^(-18t) dt
ln|y| = (5/36)C^2 e^(-18t) + Kwhere K is a constant of integration.
Taking the exponential of both sides and simplifying, we get:
y(t) = De^(5C^2/36 e^(-18t))
where D is a constant of integration.
Therefore, the general solution of the system is:
x(t) = Ce^(-9t)
y(t) = De^(5C^2/36 e^(-18t)).
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at which point (or points) on the ellipsoid x2 4y2 z2 = 9 is the tangent plane parallel to the place z = 0?
Therefore, the points on the ellipsoid where the tangent plane is parallel to the xy-plane are: (x, y, z) = (±2cosθ, sinθ, 0), z = 0 where θ is any angle between 0 and 2π.
To find the point(s) on the ellipsoid x^2/4 + y^2 + z^2/9 = 1 where the tangent plane is parallel to the xy-plane (z = 0), we need to find the gradient vector of the function F(x, y, z) = x^2/4 + y^2 + z^2/9 - 1, which represents the level surface of the ellipsoid, and determine where it is orthogonal to the normal vector of the xy-plane.
The gradient vector of F(x, y, z) is given by:
F(x, y, z) = <∂F/∂x, ∂F/∂y, ∂F/∂z> = <x/2, 2y, 2z/9>
At any point (x0, y0, z0) on the ellipsoid, the tangent plane is given by the equation:
(x - x0)/2x0 + (y - y0)/2y0 + z/9z0 = 0
Since we want the tangent plane to be parallel to the xy-plane, its normal vector must be parallel to the z-axis, which means that the coefficients of x and y in the equation above must be zero. This implies that:
(x - x0)/x0 = 0
(y - y0)/y0 = 0
Solving for x and y, we get:
x = x0
y = y0
Substituting these values into the equation of the ellipsoid, we obtain:
x0^2/4 + y0^2 + z0^2/9 = 1
which is the equation of the level surface passing through (x0, y0, z0). Therefore, the point(s) on the ellipsoid where the tangent plane is parallel to the xy-plane are the intersection points of the ellipsoid and the plane z = 0, which are given by:
x^2/4 + y^2 = 1, z = 0
This equation represents an ellipse in the xy-plane with semi-major axis 2 and semi-minor axis 1. The points on this ellipse are:
(x, y) = (±2cosθ, sinθ)
where θ is any angle between 0 and 2π.
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You Stigman notation to represent the sum of the first five terms of the following sequence -5, -9, -13
The value of Stigma notation to represent the sum of the first five terms of the sequence is -65.
The sequence can be written as : -5, -9, -13, ...
To find the sum of the first five terms of the above sequence, we need to apply the formula of Stigma notation that can be defined as the sum of the terms in a sequence. We can use Stigma notation to represent the sum of the first five terms of the given sequence. We will use the letter "n" to represent the number of terms we will use in the sequence. Then the given sequence can be represented as: S₅ = -5 - 9 - 13 - 17 - 21.To find the value of Stigma notation, the given formula can be used: Stigma notation can be defined as :∑a_n = a₁ + a₂ + a₃ + a₄ + a₅ + ... + a_n, where n represents the number of terms that are included in the sequence. To find the sum of the first five terms of the sequence: S₅ = -5 - 9 - 13 - 17 - 21We have n = 5 because we are finding the sum of the first five terms of the sequence .Therefore, ∑a₅ = -5 - 9 - 13 - 17 - 21= -65
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You win a well-known national sweepstakes. Your award is $100 a day for the rest of your life! You put the money in a bank where it earns interest at a rate directly proportional to the amount M which is in the dM account. So, =100+ KM where k is the growth constant dt m a.) Solve the DEQ (in terms of t and k) given that at t=0 days, there is no money in the account. dM 100 KM dt AM | 10/100+ KM)= t. 100+ KM = (k M= Cekt - 100 100-KM = fe at - K b.) Suppose you invest the money at 5% APR. So k=. Solve the DEQ completely. 365 c.) How much money will you have at the end of one year? d.) Assuming you live for 75 more years how much will you take to the grave with you if you never spent it? e.) How long will it take you to become a millionaire? f.) How long will it take you to become a billionaire?
a. M can be solve as M = (Ce^(kt) - 100)/K
b. The DEQ will be M = (Ce^(0.05t) - 100)/0.05
c. You will have $3,881.84 at the end of one year
d. If you live for 75 more years, you will take $13,816,540.58 to the grave with you if you never spent it
e. It will take approximately 36.23 years to become a millionaire.
f. It will take approximately 72.46 years to become a billionaire.
a) The differential equation representing the growth of the account is:
dM/dt = KM + 100
Separating the variables, we have:
dM/(KM + 100) = dt
Integrating both sides, we get:
ln(KM + 100) = kt + C
where C is the constant of integration.
Taking the exponential of both sides, we obtain:
KM + 100 = Ce^(kt)
Solving for M, we get:
M = (Ce^(kt) - 100)/K
b) Substituting k = 0.05 into the equation found in part a), we get:
M = (Ce^(0.05t) - 100)/0.05
c) To find how much money we will have at the end of one year, we can substitute t = 365 (days) into the equation found in part b):
M = (Ce^(0.05(365)) - 100)/0.05 = $3,881.84
d) Assuming we live for 75 more years, the amount of money we will take to the grave with us if we never spent it is found by substituting t = 75*365 into the equation found in part b):
M = (Ce^(0.05(75*365)) - 100)/0.05 = $13,816,540.58
e) To become a millionaire, we need to solve the equation:
1,000,000 = (Ce^(0.05t) - 100)/0.05
Multiplying both sides by 0.05 and adding 100, we get:
C e^(0.05t) = 1,050,000
Taking the natural logarithm of both sides, we obtain:
ln(C) + 0.05t = ln(1,050,000)
Solving for t, we get:
t = (ln(1,050,000) - ln(C))/0.05
We still need to find C. Substituting t = 0 and M = 0 into the equation found in part b), we get:
0 = (Ce^(0) - 100)/0.05
Solving for C, we get:
C = 5,000
Substituting this value of C into the equation for t, we get:
t = (ln(1,050,000) - ln(5,000))/0.05 ≈ 36.23 years
So it will take approximately 36.23 years to become a millionaire.
f) To become a billionaire, we need to solve the equation:
1,000,000,000 = (Ce^(0.05t) - 100)/0.05
Following the same steps as in part e), we obtain:
t = (ln(1,050,000,000) - ln(C))/0.05
Using the value of C found in part e), we get:
t = (ln(1,050,000,000) - ln(5,000))/0.05 ≈ 72.46 years
So it will take approximately 72.46 years to become a billionaire.
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test the series for convergence or divergence. [infinity] n2 8 6n n = 1
The series converges by the ratio test
How to find if series convergence or not?We can use the limit comparison test to determine the convergence or divergence of the series:
Using the comparison series [tex]1/n^2[/tex], we have:
[tex]lim [n\rightarrow \infty] (n^2/(8 + 6n)) * (1/n^2)\\= lim [n\rightarrow \infty] 1/(8/n^2 + 6) \\= 0[/tex]
Since the limit is finite and nonzero, the series converges by the limit comparison test.
Alternatively, we can use the ratio test to determine the convergence or divergence of the series:
Taking the ratio of successive terms, we have:
[tex]|(n+1)^2/(8+6(n+1))| / |n^2/(8+6n)|\\= |(n+1)^2/(8n+14)| * |(8+6n)/n^2|[/tex]
Taking the limit as n approaches infinity, we have:
[tex]lim [n\rightarrow \infty] |(n+1)^2/(8n+14)| * |(8+6n)/n^2|\\= lim [n\rightarrow \infty] ((n+1)/n)^2 * (8+6n)/(8n+14)\\= 1/4[/tex]
Since the limit is less than 1, the series converges by the ratio test.
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What is the angle of depression from the start of a 3 foot high acsess ramp that ends at a point 20 feet away along from the ground?
The angle of depression from the start of the 3-foot high access ramp to the point 20 feet away along the ground is approximately 5.71 degrees.
The angle of depression is the angle formed between the horizontal line and the line of sight to an object that is lower than the observer's level. In this case, the object is the point at the end of the access ramp, which is 3 feet high and 20 feet away from the observer. To find the angle of depression, we need to use trigonometry.
Let's call the height of the observer's eye level H, and the height of the end point of the access ramp h. We can also call the distance from the observer to the end point of the access ramp d. Using these variables, we can set up a right triangle with the vertical leg being H - h (the difference in height between the observer and the end point of the access ramp), the horizontal leg being d (the distance between the observer and the end point of the access ramp), and the hypotenuse being the line of sight from the observer to the end point of the access ramp.
Using the tangent function, we can find the angle of depression:
tanθ = opposite/adjacent = (H - h)/d
To solve for θ, we can take the inverse tangent of both sides:
θ = tan⁻¹((H - h)/d)
Assuming the observer's eye level is 5 feet above the ground, the angle of depression can be found as:
θ = tan⁻¹((5 - 3)/20) = tan⁻¹(0.1) = 5.71 degrees
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The five starters for Wolves scored all of the team’s points in the final basketball game of the season. Roger Reporter covered the game, but later his notes were accidentally destroyed. Fortunately, he had taped some interviews with the players, but when he played them back only a few quotes seemed relevant to the scoring. He knew the final score was 95-94. Using the players’ observations determine how many points each of them scored.
Player A scored 27 points, Player B scored 22 points, Player C scored 20 points, Player D scored 13 points, and Player E scored 13 points.
To determine the number of points each player scored, we can analyze the relevant quotes from the players and use the information provided. Since the final score was 95-94, the total points scored by the team must be 95.
From the quotes, we gather that Player A scored the most points, with 27. Player B mentions that Player A scored five more points than him, indicating that Player B scored 22 points.
Player C states that he scored three points less than Player B, so Player C's score is 20 points.
Player D and Player E both mention that they each scored the same number of points, and their combined total is 26 (95 - 69 = 26). Therefore, both Player D and Player E scored 13 points.
In summary, Player A scored 27 points, Player B scored 22 points, Player C scored 20 points, Player D scored 13 points, and Player E also scored 13 points, resulting in a total of 95 points for the team.
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35. ∫ 1 0 x 3 dx, using the right hand rule.
The integral ∫ 1 0 [tex]x^{3}[/tex] dx represents the area under the curve of the function y = [tex]x^{3}[/tex] from x = 0 to x = 1.
To use the right-hand rule, we divide the interval [0, 1] into n subintervals of equal width Δx, where n is a positive integer.
The right-hand rule involves approximating the area of each subinterval by the area of a rectangle with height equal to the function value at the right endpoint of the subinterval.
Therefore, the area of the nth rectangle is given by Δx * f(xn), where xn = 0 + nΔx and f(xn) = [tex](xn)^{3}[/tex]. Summing up the areas of all n rectangles gives the approximate value of the integral.
As n approaches infinity, the approximation gets closer and closer to the actual value of the integral.
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Create your own story of at least 4 sentences to explain a mnemonic phrase to help you remember the trig ratios of sine, cosine, and tangent.
Be sure your paragraph has a beginning, middle and end.
Be creative and original. Examples below.
Some Old Horse Caught Another Horse Taking Oats Away
Some Old Hippo Came A Hopping Through Our Apartment
One of the mnemonic phrases that can help to remember the trig ratios of sine, cosine, and tangent is SOHCAHTOA. This phrase stands for Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, and Tangent is Opposite over Adjacent.
To create a story that explains this phrase, we can imagine ourselves at a beach party where we meet three new friends: Sally, Carl, and Tim.Sally loves to surf, so we can remember that the Sine ratio is Opposite over Hypotenuse. Carl loves to sit on the beach and tan, so we can remember that the Cosine ratio is Adjacent over Hypotenuse. Tim loves to play beach volleyball, so we can remember that the Tangent ratio is Opposite over Adjacent. We can imagine that these three friends are holding a sign that reads "SOHCAHTOA" to help us remember the ratios. Thanks to our new friends, we'll never forget how to find the trig ratios again!This story has a beginning, middle, and end.
The beginning introduces the mnemonic phrase, the middle explains how the story can help to remember the ratios, and the end provides a conclusion by stating that the storyteller will never forget the ratios thanks to the new friends they met. This is a creative and original way to remember the trig ratios and provides an entertaining way to learn a mathematical concept.
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Explain why the following series are either convergent or divergent. No explanation yields no credit. For each series, you must state the test used, show the work related to the chosen test, and give your conclusion. (infinity) E n=1 1/(n^6 - 8)
0 ≤ 1/(n^6 - 8) ≤ 1/n^6, and ∑(n=1 to infinity) 1/n^6 converges, by the Comparison Test, we can conclude that ∑(n=1 to infinity) 1/(n^6 - 8) also converges.
To determine the convergence or divergence of the series ∑(n=1 to infinity) 1/(n^6 - 8), we can use the Comparison Test.
Comparison Test:
If 0 ≤ aₙ ≤ bₙ for all n, and ∑ bₙ converges, then ∑ aₙ also converges. Conversely, if ∑ bₙ diverges, then ∑ aₙ also diverges.
Let's analyze the given series using the Comparison Test:
Consider the series ∑(n=1 to infinity) 1/n^6.
For each term, 1/(n^6 - 8) ≤ 1/n^6 because subtracting 8 from the denominator makes it smaller.
Now, let's analyze the series ∑(n=1 to infinity) 1/n^6 using the p-series test.
p-series Test:
If ∑ 1/n^p, where p > 1, then the series converges. If p ≤ 1, the series diverges.
In our case, p = 6, which is greater than 1. Therefore, the series ∑(n=1 to infinity) 1/n^6 converges.
Since 0 ≤ 1/(n^6 - 8) ≤ 1/n^6, and ∑(n=1 to infinity) 1/n^6 converges, by the Comparison Test, we can conclude that ∑(n=1 to infinity) 1/(n^6 - 8) also converges.
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HEPLLPLLLPPPPPPOP!!!Cylinder P is transformed into cylinder Q where both the radius and height of cylinder O become double of the radius and height of cylinder P, as shown in the figure.
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Cylinder Q
If the volume of the cylinder P is 8 cubic inches, what is the volume of cylinder Q in cubic inches? Enter the answer in the box. The volume of a right circular cylinder is r? h, where r is the radius of the base of the cylinder, and h is the height of the cylinder.
The volume of cylinder Q in cubic inches will be 64 cubic inches.
Given that:
Hirght of Q, h' = 2h
Radius of Q, r' = 2r
Let r be the radius and h be the height of the cylinder. Then the volume of the cylinder will be given as,
V = π r²h cubic units
The volume of P is given as,
V = 8 cubic inches
π r²h = 8 cubic inches
The volume of Q is calculated as,
V' = π (r')²(h')
V' = π(2r)²(2h)
V' = 8πr²h
V' = 8 x 8
V' = 64 square inches
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Peanut allergies are becoming increasingly common in Western countries. Some evidence points to the timing of peanuts' first introduction in the diet as an influential factor, raising the question of whether pediatricians should recommend early exposure or avoidance. A study enrolled infants with a diagnosed peanut allergy and randomly assigned them to either completely avoid peanuts or consume peanuts in small amounts regularly until they reached 60 months of age. At the end of the study, 18 of the 51 infants who had avoided peanuts were still allergic to peanuts. In contrast, five of the 47 infants who had consumed peanuts were still allergic to peanuts.
(a) What are the two variables described? (Select two options. ) peanut allergy at 60 months of age (yes or no year the study began peanut consumption (avoid or consume small amounts) percentage of peanuts in diet choice of diaper for infants current age in months
(b) Give the marginal distribution of peanut allergy at 60 months of age, both as counts and as percentages. (Enter your answers rounded to one decimal place. ) marginal distribution (with peanut allergy, count): marginal distribution (with peanut allergy, percent): marginal distribution (no peanut allergy, count): marginal distribution (no peanut allergy, percent):
The marginal distribution of peanut allergy at 60 months of age is: With peanut allergy: 23 (23.7%) Without peanut allergy: 75 (76.3%)
(a) The two variables described are:
Peanut allergy at 60 months of age (yes or no)
Year the study began peanut consumption (avoid or consume small amounts)
(b) The marginal distribution of peanut allergy at 60 months of age, both as counts and percentages, can be calculated as follows:
Marginal distribution (with peanut allergy, count):
Number of infants still allergic to peanuts at 60 months of age = 18 + 5 = 23
Marginal distribution (with peanut allergy, percent):
Percentage of infants still allergic to peanuts at 60 months of age = (18 + 5) / (51 + 47) * 100% = 23.7%
Marginal distribution (no peanut allergy, count):
Number of infants not allergic to peanuts at 60 months of age = 51 + 47 - 23 = 75
Marginal distribution (no peanut allergy, percent):
Percentage of infants not allergic to peanuts at 60 months of age = 75 / (51 + 47) * 100% = 76.3%
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the formula i relates the coefficient of self-induction l (in henrys), the energy p stored in an electronic circuit (in joules), and the current i (in amps). find i if p and l.
The formula that relates the coefficient of self-induction l, the energy p stored in an electronic circuit, and the current i is given as: p = (1/2) * l * i^2
To find i, we can rearrange the formula as follows:
i = sqrt(2p/l)
Therefore, if we know the values of p and l, we can easily calculate the value of i using the above formula. It is important to note that the units of measurement for p, l, and i should be consistent in order for the formula to work correctly. For instance, if p is in joules and l is in henrys, then the resulting value of i will be in amps.
The energy (P) stored in an inductor with inductance (L) and current (I) is given by the formula:
P = 0.5 * L * I^2
To find the current (I) when you know the energy (P) and inductance (L), you can rearrange the formula as follows:
I = sqrt(2 * P / L)
This equation allows you to calculate the current in the electronic circuit given the energy stored in the inductor and the inductor's coefficient of self-induction.
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Suppose that f(x)>0 on [-2,5] is a continuous function. then the area beneath the curve f(x) on [-2,5] is given by:∫ f(x) dx
The area beneath the curve f(x) on [-2,5] is given by the integral: ∫[-2,5] f(x) dx.
To find the area, follow these steps:
1. Identify the given function f(x), which is continuous and positive on the interval [-2, 5].
2. Determine the limits of integration, which are -2 (lower limit) and 5 (upper limit).
3. Integrate the function f(x) with respect to x from -2 to 5.
4. Evaluate the definite integral, which will give you the area beneath the curve.
The area represents the accumulated value of the function f(x) over the specified interval, considering its positive values on the interval [-2, 5].
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