if y is given and you need to find third derivative of y (given as y'''), what are the steps: explain what one needs to do and say it in your words.

Therefore , the solution of the given problem of unitary method comes out to be common shares of Zeil Inc. is 2.23%.

This common convenience, already-existing variables, or all important elements from the original Diocesan adaptable study that followed a particular methodology can all be used to achieve the goal. Both of the crucial elements of a term affirmation outcome will surely be missed if it doesn't happen, but if it does, there will be another chance to get in touch with the entity.

Here,

Earnings per Share are calculated as (Net Income – Preferred Dividends) / the average number of outstanding Common Shares.

=>  Market price per share / earnings per share is the Price-Earnings Ratio.

=> Dividends per Share are calculated as follows: Common Stock Dividends / Average Common Shares Outstanding

=>  Dividend Yield is the product of dividends per share and the share price.

=>  (Beginning Common Shares plus Ending Common Shares) / 2 equals the average number of Common Shares Outstanding.

=>  Starting common shares equals ending common shares, which is

=>  $3,500,000 / $25, or 140,000.

(a) The earnings per share are ($424,000 - $0) / 140,000, which equals $3.03.

The ordinary stock price of Zeil Inc.

(b) The price-earnings ratio for Zeil Inc.'s common shares is 11.20 divided by 3.03, or 3.69.

(c) Dividends per Share: $35,000./140,000. = $0.25

Therefore, $0.25 in dividends are paid per unit of Zeil Inc. common stock.

(d) Dividend Yield: $0.25 divided by $11.20 equals 0.0223, or 2.23%.

The common shares of Zeil Inc. is 2.23%.

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If Andres Michael bought a new boat. He took out a loan for $24,420 at 3.5% interest for 2 years. Andres Michael owes $18806.6 at maturity.

To calculate how much is due at maturity, we first need to determine how much of the loan remains after the two partial payments.

To do this, we can use the formula for simple interest:

I = P * r * t

Where:

I = Interest

P = Principal (original loan amount)

r = Annual interest rate

t = Time (in years)

The interest for the first two months can be calculated as:

I1 = P * r * t1

= 24420 * 0.035 * (2/12)

= 142.45

So after the first two months, the amount owing on the loan is:

P1 = P + I1 - 4330

= 24420 +142.45 - 4330

= 20,232.45

The interest for the next four months can be calculated as:

I2 = P1 * r * t2

= 20,232.45 * 0.035 * (4/12)

= 236.05

So after six months, the amount owing on the loan is:

P2 = P1 + I2 - 2600

=  20,232.45 + 236.05- 2600

= 17868.50

Now we can calculate the interest for the remaining 18 months:

I3 = P2 * r * t3

=  17868.50* 0.035 * (18/12)

= 938.10

So the total amount owing at maturity (after 2 years) is:

Total amount owing = P2 + I3

=  17868.50 + 938.10

= 18806.6

Therefore, Andres Michael owes $18806.6 at maturity.

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Elliptic curve cryptography (ECC) makes use of elliptic curves in which the variables and coefficients are all restricted to elements of a finite field.

ECC is a type of public-key cryptography that is based on the difficulty of solving the elliptic curve discrete logarithm problem (ECDLP), which is a variant of the discrete logarithm problem in which the group operation is performed on points on an elliptic curve.

ECC is particularly useful in settings where computational resources are limited, such as mobile devices and smart cards, as it provides the same level of security as other public-key cryptographic systems but with smaller key sizes.

ECC also offers other advantages over traditional public-key cryptography such as faster computation times, lower power consumption, and smaller message sizes.

ECC is widely used in a variety of applications, including digital signatures, encryption, and key exchange. It is implemented in many cryptographic standards, such as the Transport Layer Security (TLS) protocol used to secure internet communications, and is considered to be one of the most promising cryptographic techniques for the future.

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Complete question is:

___________ makes use of elliptic curves in which the variables and coefficients are all restricted to elements of a finite field.

(a) The function of price is $1600

(b) The marginal revenue is $80

(c) The revenue and marginal revenue when price is $5 is $30

Marginal revenue is the increase in revenue from selling an additional unit of product. Although marginal revenue may remain constant at a certain level of output, it follows the law of diminishing returns and eventually declines as output levels increase. In economic theory, perfectly competitive firms continue to produce until marginal revenue equals marginal cost.

Assume that a certain product has the demand function given by:

9 = 1000e -0.02p

R(x) = 80x

P(x) = -0.25x² + 40x -100

R'(x) =80

P'(x) = -0.5x + 40

Because we have refurbished x = 20 iPad this month x = 20.

Thus,

R(20) = 80(20) = $1600

P(20) = -0.5(20)² + 40(20) - 1000

         = -$300

R'(20)= $80

And, P'(20) = -0.5 (20) + 40

                   = $30

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Answer:

$586.00

Step-by-step explanation:

Markup is the how much more an item or service is sold for to cover overhead fees. If the markup is 25%, then the price was increased by 25% in order to be sold for $732.50. We can set up a proportion to represent this where c is the cost.

[tex]\frac{732.5}{1.25} = \frac{c}{1.00}[/tex]

Cross-multiply.

1.25c = 732.5

c = 586

So, the cost of the item was $586.00

Check the picture below.

so if the triangles are congruent, then CPCTC.

If x and y are integer variables, then the expression that evaluates to true if x is greater than y is "x>y".

In Java, symbol of ">" is used for "greater-than" operator. So, the expression which evaluates to "true" if integer "x" is greater than integer "y" is "x > y".

This expression compares the values of x and y and returns a Boolean value of "true" if x is greater than y, and "false" otherwise.

The expression can be used in conditional statements, loops, and other constructs that require a Boolean value as a condition. It is important to note that the ">" operator only works with primitive types such as int, long, double, etc.

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This means that we can be 95% confident that the true difference in means between Clarion and Wabash is somewhere between -3.33 and -0.88

A 95% confidence interval for the difference in means between Clarion and Wabash can be calculated using the following formula: CI95 = (μ1 - μ2) ± 1.96*√(σ1^2/n1 + σ2^2/n2),where μ1 and μ2 are the population means of Clarion and Wabash respectively, σ1 and σ2 are the population standard deviations of Clarion and Wabash respectively, and n1 and n2 are the sample sizes of Clarion and Wabash respectively. To calculate the confidence interval, we need to have access to the population means and standard deviations of Clarion and Wabash, which we do not have. In their place, we can use the sample means and standard deviations as an estimate of the population means and standard deviations. Using the sample means and standard deviations, the 95% confidence interval for the difference in means between Clarion and Wabash is (-3.33, -0.88). This means that we can be 95% confident that the true difference in means between Clarion and Wabash is somewhere between -3.33 and -0.88.

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What is the approximate 95% confidence interval for the difference in means between Clarion and Wabash?

Step-by-step explanation:

Please mark as brainliest

Answer:

We can only see g(x) not f(x)

Step-by-step explanation:

Domain of g(x) is

[tex]( - \infty \: to \: \infty )[/tex]

Range of g(x) is

[tex](0 \: to \: \infty )[/tex]

Range lf

The joint probability density function for the random variables Y1 and Y2 E(Y1+Y2) and V(Y1+Y2) is 41/144.

To find E(Y1+Y2), we need to integrate the sum of Y1 and Y2 over their joint probability density function:

E(Y1+Y2) = ∫∫ (y1 + y2) f(y1,y2) dy1 dy2

= ∫∫ (y1 + y2) (2) dy1 dy2, where the limits of integration are 0 to 1 for both y1 and y2 and y1+y2 <=1

= ∫[tex]0^1[/tex] ∫[tex]0^{(1-y1)}[/tex](y1 + y2) (2) dy2 dy1

= ∫[tex]0^1[/tex] (2y1 + 1) (1-y1)² dy1

= 5/12

To find V(Y1+Y2), we can use the formula V(Y1+Y2) = E[(Y1+Y2)²] - [E(Y1+Y2)]².

First, we need to find E[(Y1+Y2)^2]:

E[(Y1+Y2)²] = ∫∫ (y1+y2)² f(y1,y2) dy1 dy2

= ∫∫ (y1² + y2² + 2y1y2) (2) dy1 dy2, where the limits of integration are 0 to 1 for both y1 and y2 and y1+y2 = 1

= ∫[tex]0^1[/tex] ∫[tex]0^{(1-y1)}[/tex] (y1² + y2² + 2y1y2) (2) dy2 dy1

= ∫[tex]0^1[/tex] (1/3)y1³ + (1/2)y1² + (1/2)y1

(1/3)y1 + (1/4) dy1

= 7/12

Next, we need to find [E(Y1+Y2)]²:

[E(Y1+Y2)]² = (5/12)² = 25/144

Therefore, V(Y1+Y2) = E[(Y1+Y2)²] - [E(Y1+Y2)]² = (7/12) - (25/144) = 41/144.

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Toward the end of a game of Scrabble, you hold the letters D, O, G, and Q. You can choose 3 of these 4 letters and arrange them in order in 24 different ways.

To solve this problem, we need to use the concept of permutations. A permutation is an arrangement of objects in a specific order. In this case, we need to find the number of permutations that can be made from the letters D, O, G, and Q when we choose 3 of these 4 letters.

The formula for finding the number of permutations is:

n! / (n-r)!

where n is the total number of objects and r is the number of objects we choose.

Using this formula, we can calculate the number of permutations as follows:

4! / (4-3)!

= 4! / 1!

= 4 x 3 x 2 x 1 / 1

= 24

Therefore, we can arrange the chosen 3 letters in 24 different ways.

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The probability that the sample mean of ln(LC50) will differ from the true population mean by no more than 0.5 is 0.0019.

The concept of independent random variables is very similar to independent events. Recall that two events A and B are independent if we have P(A,B) = P(A)P(B) and remember that comma means sum, i.e.

         P(A, B)=P( A and B)=P (A∩B).

Similarly, we have the following definition for independent discrete random variables.

Assuming 'X' and 'Y' are independent random samples

X : mean value =  , variance = 0.4 /10 = 0.04

Y : mean value =  , variance = 0.8/10 = 0.08

since the values are independent

V [ X - Y ] = V [ X ] + V [Y ] = 0.04 + 0.08 = 0.12

Now,

The probability that the sample mean for species A exceeds the sample mean for species B by at least 1 unit

  [tex]P(X-Y\geq 1) = P{ \frac{(X-Y)-(U_1-U_2)}{\sqrt{V(X-Y} } \geq \frac{1-0}{\sqrt{0.12} }[/tex]

⇒ P{Z ≥ 2.8858}

⇒ 1 -P{Z≤ 2.8858} = 1 - 0.9981

⇒ P{X-Y≥ 1} = 0.0019.

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Therefore, the monthly payment for a student loan of $15,000 at a fixed APR of 12% for 20 years is $144.36.

Interest is the cost of borrowing money or the return on investing money. When you borrow money, you usually have to pay back more than you borrowed, and the additional amount you pay is the interest. The interest rate is expressed as a percentage of the borrowed amount, and it can vary depending on factors such as the borrower's credit score, the term of the loan, and the lender's policies.

Given by the question.

Assuming the loan has a fixed interest rate of 12% per annum, the amount of interest charged each year will be:

12% of $15,000 = $1,800

The total interest charged over 20 years will be:

$1,800 x 20 = $36,000

The total amount to be repaid (principal + interest) will be:

$15,000 + $36,000 = $51,000

If the loan is being repaid in equal monthly installments over the 20-year term, the monthly payment can be calculated using the following formula:

M = P * (r[tex](1+r)^{n}[/tex]) / ([tex](1+r)^{n}[/tex]- 1)

Where:

M = Monthly payment

P = Principal amount (in this case, $15,000)

r = Monthly interest rate (12% per annum / 12 months = 1% per month)

n = Total number of payments (20 years x 12 months per year = 240)

Plugging in the values:

M = $15,000 * (0.01[tex](1+0.01)^{240}[/tex]) / ([tex](1+0.01)^{240}[/tex] - 1)

M = $144.36

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The area to the right of x is 0.877.

In a normal distribution, the entire area under the curve is identical to 1. The area to the left of a specific value of x represents the possibility of observing a value largely lesser than or same tox.

However, we're capable to discover the area to the right of x with the aid of abating the left area from 1, If the place to the left of x is given.

In this case, the area to the left of x is 0.123. thus, the place to the right of x is

1-0.123 = 0.877

Thus, the area is 0.877 to the right of x.

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Based on the differential equation and the initial condition, the population of the mouse species never becomes zero. Therefore, there is no time at which the population becomes zero.

We can begin by separating variables and integrating both sides of the equation

dt/dp(t) = 21/p(t) - 450

dt = (1/21) * (1/p(t) - 450) dp(t)

Integrating both sides gives

t + C = (1/21) * ln|p(t)| + 450t + D

where C and D are constants of integration. We can solve for these constants using the initial condition p(0) = 850

0 + C = (1/21) * ln|850| + 0 + D

C = (1/21) * ln|850| - D

We can simplify this expression by defining a new constant E = (1/21) * ln|850| - D

C = E - D

Substituting this expression for C back into our previous equation, we have

t + E - D = (1/21) * ln|p(t)| + 450t

Solving for p(t), we get

ln|p(t)| = 21(450t + D - E) + ln|850|

p(t) = ± e^(21(450t + D - E) + ln|850|)

Since p(t) represents a population, we can discard the negative solution and take only the positive solution

p(t) = e^(21(450t + D - E) + ln|850|)

We want to find the time at which the population becomes zero, so we set p(t) = 0 and solve for t

0 = e^(21(450t + D - E) + ln|850|)

ln|0| = 21(450t + D - E) + ln|850|

This is not possible, since ln|0| is undefined. Therefore, the population never becomes zero.

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The slope of this linear function is equal to: B. -2/9.

The volume of a cylinder with a height of 10 m and a radius of 5 m is equal to 785 m³.

The value of each expression is: C. a) 2, b) 1/2, c) 2/9.

In Mathematics, the slope of any straight line can be determined by using the following mathematical equation;

Slope (m) = (Change in y-axis, Δy)/(Change in x-axis, Δx)

Slope (m) = (y₂ - y₁)/(x₂ - x₁)

Slope (m) = (8 - 10)/(6 - (-3))

Slope (m) = (8 - 10)/(6 + 3)

Slope (m) =

Slope (m) = -2/9.

In Mathematics, the volume of a cylinder can be calculated by using this formula:

Volume of a cylinder, V = πr²h

Where:

By substituting the given parameters, we have:

Volume of cylinder, V = 3.14 × 5² × 10

Volume of cylinder, V = 785 m³

(√2)² = 2

(1/√2)² = 1/2

(√2/3)² = 2/9

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Answer:

36.3636364%

or 36.36

Step-by-step explanation:

Answer:

Let's work backwards from the end of lesson 4 to figure out how many sweets were left in Anna's bag after each lesson. We know that she had 1 sweet left at the end of lesson 4, so before that she must have had:

Lesson 4: 1 sweet + 1 sweet for teacher + 1 sweet left over = 3 sweets

Lesson 3: (3 sweets + 1 sweet for teacher) x 2 = 8 sweets

Lesson 2: (8 sweets + 1 sweet for teacher) x 2 = 18 sweets

Lesson 1: (18 sweets + 1 sweet for teacher) x 2 = 40 sweets

So, Anna started with 40 sweets in her bag.

The value of the x is 5√3 after we successfully do the application of the 30°-60°-90° Triangle theorem.

The 30°-60°-90° Triangle Theorem states that in such a triangle, the side opposite the 30° angle is half the length of the hypotenuse, and the side opposite the 60° angle is the product of the length of the hypotenuse and the square root of 3 divided by 2.

Using this theorem, we can write:

y = hypotenuse

Opposite of 30° angle = 5 = hypotenuse/2

Opposite of 60° angle = x = hypotenuse × (√(3)/2)

Solving for the hypotenuse in terms of y from the first equation, we get:

hypotenuse = 5×2 = 10

Substituting this value into the third equation, we get:

x = 10 × (√(3)/2) = 5 × √(3)

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Answer:

To find the probability that a randomly selected person is a meat-eater, we need to add up the number of meat-eaters and divide by the total number of individuals surveyed. From the given table, we can see that there are 72 meat-eaters out of a total of 190 individuals surveyed:

Total meat-eater = 72

Total surveyed = 190

So the probability of selecting a meat-eater is:

P(meat-eater) = Total meat-eater / Total surveyed

P(meat-eater) = 72 / 190

P(meat-eater) = 0.38 (rounded to the hundredths place)

Therefore, the probability that a randomly selected person is a meat-eater is 0.38 or 38%.

By answering the question the answer is standard deviation  For 0 courses: 3 students (3/55 ≈ 0.1 or 9.1%); For 1 course: 12 students (12/55 ≈ 0.2 or 21.8%)

Standard deviation is a statistic that describes the variability or variance of a group of numbers. A high standard deviation indicates that the values ​​are more dispersed, while a low standard deviation indicates that the values ​​tend to be closer to the established mean. A measure of how far the data are from the mean is the standard deviation (or ). If the standard deviation is small, the data tend to be clustered around the mean, and if the standard deviation is large, the data are more dispersed. The average variability of the dataset is measured as standard deviation. Shows the mean deviation of each score from the mean. 

To fill in the blank cells, we need to calculate the number of students who reported each course number and the relative frequency (rounded to the nearest tenth). This can be done like this:

For 0 courses:

3 students (3/55 ≈ 0.1 or 9.1%)

For 1 course:

12 students (12/55 ≈ 0.2 or 21.8%)

For 2 courses:

17 students (17/55 ≈ 0.3 or 30.9%)

For 3 courses:

9 students (9/55 ≈ 0.2 or 16.4%)

For 4 courses:

8 students (8/55 ≈ 0.1 or 14.5%)

For 5 courses:

2 students (2/55 ≈ 0.0 or 3.6%)

For 6 courses:

4 students (4/55 ≈ 0.1 or 7.3%)

The finished table looks like this:

+--------+--------+---------------------+

| Number | Number | Relative Frequency  |

|  of    | of     | (Rounded to nearest |

|Courses |Students|         tenth)      |

+--------+--------+---------------------+

|   0    |   3    |         0.1         |

|   1    |   12   |         0.2         |

|   2    |   17   |         0.3         |

|   3    |   9    |         0.2         |

|   4    |   8    |         0.1         |

|   5    |   2    |         0.0         |

|   6    |   4    |         0.1         |

+--------+--------+---------------------+

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The equation of the line passing through (-3, -1) with slope 3/5 is y = (3/5)x + 4/5.

What is point slope form?

The equation of a line is expressed in the point-slope form as follows: y - y1 = m(x - x1), where m is the slope of the line and (x1, y1) is a point on the line. When we know the slope of a line and a point on the line but not the intercepts, this version of the equation is helpful. It eliminates the need to independently compute the intercepts by allowing us to state the equation of the line in terms of the given point and slope.

Given that, line passes through (-3, -1) and has a slope of 3/5.

The points slope form is given as:

y - y1 = m(x - x1)

Substituting the values we have:

y - (-1) = (3/5)(x - (-3))

y + 1 = (3/5)x + 9/5

y = (3/5)x + 4/5

Therefore, the equation of the line passing through (-3, -1) with slope 3/5 is y = (3/5)x + 4/5.

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From the given probability distribution, the probability of x being 4 in the given probability distribution is 0.15,

According to the given probability distribution in Table 1, the probability of x being 4 is 0.15. This means that out of all the possible values of x (0 to 6), there is a 15% chance that x will be equal to 4.

To understand the probability distribution better, we can visualize it using a graph. The x-axis represents the possible values of x, while the y-axis represents the probability of each value. We can plot the values from Table 1 to create a histogram or a bar graph.

From the graph, we can see that the probability distribution is skewed to the right, with the highest probability being at x=2. This means that there is a higher chance that x will be closer to 2 than to 0 or 6.

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The given statement " show that there exist coefficients w0, w1, ..., wn that depend on x0, x1, ..., xn, and on a and b, such that the limit of the sum, as a approaches b, of the form summation from i=0 to n of wi*p(xi) for all polynomials p of degree <= n", is proved by the use of Lagrange form of the interpolating polynomials.

Let p(x) be a polynomial of degree at most n. Then, by the Lagrange interpolation formula from Section 4.1, we have:

p(x) = Summation from i=0 to n of p(xi) * Li(x)

where Li(x) is the ith Lagrange basis polynomial, defined by:

Li(x) = Product from j=0 to n, j != i, of (x - xj) / (xi - xj)

Now, consider the sum:

S = Summation from i=0 to n of wi * p(xi)

where wi are coefficients to be determined. We want to show that the limit of S as a approaches b exists for all polynomials p of degree at most n.

We can express S in terms of the Lagrange basis polynomials as:

S = Summation from i=0 to n of wi * p(xi)

= Summation from i=0 to n of wi * Summation from j=0 to n of p(xj) * Li(xj)

= Summation from j=0 to n of p(xj) * Summation from i=0 to n of wi * Li(xj)

Note that the summation over i is only dependent on the Lagrange basis polynomial Li(xj), and does not depend on p(xj). Therefore, we can choose the coefficients wi such that:

Summation from i=0 to n of wi * Li(xj) = 0 for j != k

Summation from i=0 to n of wi * Li(xk) = 1

for some k in {0, 1, ..., n}.

To see why this is possible, note that the Lagrange basis polynomials satisfy the property that Li(xi) = 1 and Li(xj) = 0 for j != i. Therefore, we can choose the coefficients wi to be:

wi = Li(xk) / Summation from i=0 to n of Li(xk)

which gives:

Summation from i=0 to n of wi * Li(xj) = Li(xk) / Summation from i=0 to n of Li(xk) * Summation from i=0 to n, i != k of Li(xj)

= 0 for j != k

Summation from i=0 to n of wi * Li(xk) = 1

Now, we have:

S = Summation from j=0 to n of p(xj) * Summation from i=0 to n of wi * Li(xj)

= Summation from j=0 to n of p(xj) * Li(xk)

Taking the limit as a approaches b, we get:

lim a->b S = lim a->b Summation from j=0 to n of p(xj) * Li(xk)

= Summation from j=0 to n of p(xj) * lim a->b Li(xk)

= Summation from j=0 to n of p(xj) * Integral from a to b of Li(x) dx

where we have used the fact that the limit and integral commute, and the limit of the Lagrange basis polynomial Li(xk) is equal to the integral of Li(x) over the interval [a, b], which is a constant that does not depend on k.

Therefore, we have shown that there exist coefficients w0, w1, ..., wn that depend on x0, x1, ..., xn, and on a and b, such that the limit of the sum, as a approaches b, of the form Summation from n to i=0 wi p(xi).

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_____The given question is incomplete, the complete question is given below:

Show that there exist coefficients w0,w1, . . . ,wn depending on x0, x1, . . . , xn and on a, b such that limit a to b { summation n to i=0 wi p(xi)} for all polynomials p of degree ?n.

   Hint: Use the Lagrange form of the interpolating polynomials from Section 4.1

Answer:

Proofs attached to answer

Step-by-step explanation:

Proofs attached to answer

Answer:

19/20 , 1 1/4 , 1 4/5

Step-by-step explanation:

1.8 = 1 4/5 (fraction)

1.8 converts to 18/10. This can be simplified twice, firstly by making it 9/5 since both 18 and 10 are divisible by two, but can be simplified further to 1 4/5

1.25 = 1 1/4 (fraction)

1.25 converts to 125/100. This can be simplified to 5/4 or 1 1/4

0.95 = 19/20 (fraction)

0.95 converts to 95/100. This can be simplified to 19/20

Ascending Order (smallest to largest)

smallest - 19/20

middle - 1 1/4

largest - 1 4/5

I believe this is the right answer, but haven't done fractions in a while so may want to double check to make sure

We can conclude that Maya's experimental probabilities fluctuate around the theoretical probability, but over a larger number of trials, the experimental probabilities should converge towards the theoretical probability.

What is probability?

Probability is simply how likely something is to happen. Whenever we're unsure about the outcome of an event, we can talk about the probabilities of certain outcomes—how likely they are. The analysis of events governed by probability is called statistics.

The theoretical probability of rolling a 5 on a fair die is 1/6, which means that if the die is rolled many times, we would expect to see a 5 about 1/6 of the time.

For the first 100 trials, Maya rolled a 5 on 25 of those trials. The experimental probability of rolling a 5 in this case is:

experimental probability = number of 5's rolled / number of trials

experimental probability = 25/100

experimental probability = 0.25

So, in the first 100 trials, Maya's experimental probability of rolling a 5 was 0.25.

For the first 200 trials, Maya rolled a 5 on 30 of those trials. The experimental probability of rolling a 5 in this case is:

experimental probability = number of 5's rolled / number of trials

experimental probability = 30/200

experimental probability = 0.15

So, in the first 200 trials, Maya's experimental probability of rolling a 5 was 0.15.

Comparing these experimental probabilities to the theoretical probability, we see that after 100 trials, Maya's experimental probability of rolling a 5 (0.25) is higher than the theoretical probability (1/6 ≈ 0.167). This suggests that Maya's sample of 100 trials was somewhat biased in favor of rolling a 5.

On the other hand, after 200 trials, Maya's experimental probability of rolling a 5 (0.15) is lower than the theoretical probability (1/6 ≈ 0.167). This suggests that Maya's sample of 200 trials was somewhat biased against rolling a 5.

Overall, we can conclude that Maya's experimental probabilities fluctuate around the theoretical probability, but over a larger number of trials, the experimental probabilities should converge towards the theoretical probability. This is known as the law of large numbers, which states that as the number of trials or observations increases, the experimental probability will tend to approach the theoretical probability.

Learn more about probability on:

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We might say that Maya's experimental probabilities oscillate about the theoretical probability, but after more trials, the experimental probabilities ought to converge to the theoretical probability.

What is probability?

Simply put, probability is the likelihood that something will occur. When we don't know how an event will turn out, we can discuss the likelihood or likelihood of several outcomes. Statistics is the study of events that follow a probability distribution.

A fair die has a theoretical probability of rolling a 5 of 1/6, therefore if the die is rolled several times, we can anticipate seeing a 5 roughly 1/6 of the time.

For the first 100 trials, Maya rolled a 5 on 25 of those trials. The experimental probability of rolling a 5 in this case is:

experimental probability = number of 5's rolled / number of trials

experimental probability = 25/100

experimental probability = 0.25

So, in the first 100 trials, Maya's experimental probability of rolling a 5 was 0.25.

For the first 200 trials, Maya rolled a 5 on 30 of those trials. The experimental probability of rolling a 5 in this case is:

experimental probability = number of 5's rolled / number of trials

experimental probability = 30/200

experimental probability = 0.15

So, in the first 200 trials, Maya's experimental probability of rolling a 5 was 0.15.

Comparing these experimental probabilities to the theoretical probability, we see that after 100 trials, Maya's experimental probability of rolling a 5 (0.25) is higher than the theoretical probability (1/6 ≈ 0.167). This suggests that Maya's sample of 100 trials was somewhat biased in favor of rolling a 5.

On the other hand, after 200 trials, Maya's experimental probability of rolling a 5 (0.15) is lower than the theoretical probability (1/6 ≈ 0.167). This suggests that Maya's sample of 200 trials was somewhat biased against rolling a 5.

Overall, we can conclude that Maya's experimental probabilities fluctuate around the theoretical probability, but over a larger number of trials, the experimental probabilities should converge towards the theoretical probability. This is known as the law of large numbers, which states that as the number of trials or observations increases, the experimental probability will tend to approach the theoretical probability.

Learn more about probability on:

https://brainly.com/question/13604758

#SPJ1

Answer:

huh

Step-by-step explanation: