David is working two summer jobs, making $13 per hour landscaping and making $8 per hour clearing tables. In a given week, he can work no more than 16 total hours and must earn no less than $160. Also, he must work at most 13 hours landscaping. If
� x represents the number of hours landscaping and �y represents the number of hours clearing tables, write and solve a system of inequalities graphically and determine one possible solution.

The average square distance from the origin to a point in the domain represented by the given figure is approximately 9.593 units squared.

We can use the equation of the ellipse to solve for y in terms of x. Specifically, we can rearrange the equation as:

y² = b²(1 - x²/a²)

Taking the square root of both sides, we get:

y = ±sqrt[b²(1 - x²/a²)]

Since we only need to consider the part of the domain where y is positive, we can use the positive square root:

y = sqrt[b²(1 - x²/a²)]

Now we can calculate the square distance from the origin to a point (x, y) in the domain. This is given by:

d² = x² + y²

Substituting our expression for y, we get:

d² = x² + b²(1 - x²/a²)

Simplifying, we get:

d² = b² - (b² - a²)x²/a²

Now we can integrate this expression over the interval [0, a] to find the total sum of square distances. Specifically, we have:

∫[0,a] (b² - (b² - a²)x²/a²) dx

This integral can be evaluated using basic calculus techniques. After integrating and simplifying, we get:

(b²/2)(1 - (1/2)(b/a)²)

Finally, to find the average square distance, we divide this sum by the total number of points in the domain, which is given by the area of the quarter-ellipse. This area can be found using basic geometry, and is:

(1/4)πab

Dividing the sum by the area, we get the average square distance:

(b²/2)(1 - (1/2)(b/a)²) / [(1/4)πab]

Simplifying, we get:

(4b²/πa)(1/4 - (1/2)(b/a)²)

Substituting in the given values of a and b, we get the final answer:

(4(12²/π(6))(1/4 - (1/2)(12/6)²)

Simplifying further, we get:

(48/π)(1/4 - 1)

= -9.593

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(a) Probability that X < 0 is P(X<0) = 1/2
(b) Probability that -3.5 < X < 3.5 is given by = 1/2
(c) Probability that -5 ≤ X ≤ 6 is given by = 11/14.
(d) Let k be any number such that -7 < k < k+4 < 7 = 2/7

(a) Since the distribution is uniform, the probability of X being less than 0 is equal to the proportion of the interval (-7, 7) that lies to the left of 0. This proportion is (0 - (-7))/(7 - (-7)) ⇒ 7/14 ⇒ 1/2.

Therefore, P(X < 0) = 1/2.

(b) Following the same logic as above, the probability of X lying between -3.5 and 3.5 is equal to the proportion of the interval (-7, 7) that lies between -3.5 and 3.5. This proportion is (3.5 - (-3.5))/(7 - (-7)) ⇒ 7/14 ⇒ 1/2.

Therefore, P(-3.5 < X < 3.5) = 1/2.

(c) Similarly, the probability of X lying between -5 and 6 is equal to the proportion of the interval (-7, 7) that lies between -5 and 6. This proportion is (6 - (-5))/(7 - (-7)) ⇒ 11/14.

Therefore, P(-5 ≤ X ≤ 6) = 11/14.

(d) The interval (k, k+4) lies completely within the interval (-7, 7) if -3 < k < 3. If k satisfies this inequality, then the probability of X lying between k and k+4 is equal to the proportion of the interval (-7, 7) that lies between k and k+4, which is (k+4 - k)/(7 - (-7)) ⇒ 4/14 ⇒ 2/7.

Therefore, P(k < X < k+4) = 2/7.

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The value of lim f(x)g(x) as x approaches infinity is 0.

L'Hopital's rule is a mathematical tool used to evaluate limits of functions that are in an indeterminate form.

To find the limit of f(x)g(x) as x approaches infinity, we can use L'Hopital's rule since it is an indeterminate form of infinity times zero. We have:

lim x→[infinity] f(x)g(x) = lim x→[infinity] [(3x + 2e^(-3x))(1/x + 4)]

= lim x→[infinity] [(3 + 2e^(-3x)/x)/(1/x + 4)^(-1)]

Applying L'Hopital's rule to the fraction in the numerator, we get:

lim x→[infinity] [(2e^(-3x)(-3)/x^2)/(1/x + 4)^(-1)]

= lim x→[infinity] [(6e^(-3x)/x^2)/(1/x + 4)]

= lim x→[infinity] [(6e^(-3x)/(x + 4x^2))]

= 0

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Answer:

7x+4y

Step-by-step explanation:

combine like terms

Answer:

4?

Step-by-step explanation:

It is a composite number, its proper divisors being 1, 2, 4, 7, and 14.

Answer:

-3+[tex]4x^{2}[/tex]

Step-by-step explanation:

The weight of 12 badminton rackets, each weighing 0.950 kg, is 11.4 kg.

The weight of an object refers to the force exerted by gravity on that object. This force is typically measured in units of mass, such as kilograms or pounds. In the case of badminton rackets, the weight is usually measured in grams.

Now, let's consider the weight of a single badminton racket. You've been given the information that the weight of one racket is 0.950 kg. This means that the force exerted by gravity on that racket is 0.950 kg.

If you want to find the weight of 12 such rackets, you need to multiply the weight of one racket by 12. This gives you:

0.950 kg x 12 = 11.4 kg

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Answer:

2/5

Step-by-step explanation:

3/5 * 2/3 is your equation

Multiply the numerators: 3*2=6

Multiply the denominators: 5*3=15

You get: 6/15

Simplify by dividing both the numerator and denominator by 3.

6/15=2/5

The multiplicity of a root r of the characteristic equation of A is called the algebraic multiplicity of r as an eigenvalue of A.

The above statement is True.

Eigenvalue:

An eigenvalue is a special set of scalar values ​​associated with the most probable system of linear equations in a matrix equation. Eigenvectors are also called eigenvalues. It is a non-zero vector which can be modified by at most its scalar factor after applying a linear transformation.

According to the Question:

If the geometric multiple of the eigenvalues ​​is greater than or equal to 2, the linearly independent set of eigenvectors is no longer unique to the multiple as before. For example, for the diagonal matrix A=[3003], one could also choose the eigenvectors [11] and [1−1], or any pair of two linearly independent vectors.

Sometimes vectors are simply expanded to vector times matrix. If this happens, this vector is called the eigenvector of the matrix and the "stretch factor" is called the eigenvalue. Example: Given a square matrix A, λ is the eigenvalue of A, and the corresponding eigenvector x is

Ax = λx.

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Answer:

0

Step-by-step explanation:

0

Thank me...........

Answer:

The three angles of a triangle are, ∠PQR, ∠QRP, and ∠RPQ respectively. The three sides of a triangle are side PQ, side QR, and side RP

Step-by-step explanation:

please brainliest!!

The answer of the given question based on the finding the m∠QTS from a circle with the measure of minor arc the answer is the measure of angle ∠QTS is 60° degrees.

In geometry, arc is  portion of a curve or  circle. It is defined as the part of the curve that is contained between two points, called endpoints, on curve. The length of  arc is  fraction of the circumference of  circle, determined by the measure of  central angle that subtends it.

An arc is typically denoted by  two endpoints, with  small arc or arc symbol above the two letters. For example, an arc with endpoints A and B would be denoted as "⌒AB" or "AB".Arcs can be classified based on their measure or  position of their endpoints.

An arc with measure less than 180 degrees is called  minor arc, while an arc with measure greater than 180 degrees is called  major arc. An arc with measure equal to 180 degrees is called semicircle.

Since arc QS is a minor arc, we know that angle ∠QTS is half the measure of arc QS. Therefore, we have:

m∠QTS = (1/2) m arc QS

m∠QTS = (1/2) (120°) [Substituting the given value of m arc QS]

m∠QTS = 60°

Therefore, the measure of angle ∠QTS is 60° degrees.

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The correct option is D i.e.35. A line segment connecting the rectangle's two opposed vertices is known as the diagonal.

The rectangle is split into two identical right triangles by its diagonal. The two-dimensional shape of a rectangle has four sides, four vertices, and four angles. A rectangle's two diagonals are the same length.

Two diagonals divide a rectangle into two right-angled triangles, with the diagonal serving as the hypotenuse on each diagonal. The diagonals cut each other in half, creating an acute angle and an obtuse angle respectively. The diagonal of the rectangle form two congruent triangles.

Since the diagonals of a rectangle are equal

∴ QS = RT

20x - 5 = 14x + 7

6x = 12

x = 2

QS = RT = 35

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The radius of the given circle is x = 3 and x = 3/11.

A polynomial equation of degree two is a quadratic equation, indicating that the variable's maximum exponent is 2. There are various ways to solve a quadratic equation, including factoring, completing the square, and applying the quadratic formula. Finding two quadratic expression factors that multiply to produce the constant term c and add to produce the linear term b's coefficient is known as factoring. The quadratic expression is completed by adding and removing a constant component to create a perfect square trinomial that can be factored or solved by calculating the square root.

In the given figure connect the points QF and QG using a line segment, thus forming two right angled triangle.

The perpendicular segment QA divides the base into two equal parts. Thus, the base of the triangle is 8.

Using the Pythagorean theorem we have:

(QF)² = (8)² + (4x + 3)²

QG² = 8² + (7x - 6)²

In the figure QF = QG = r thus:

(8)² + (4x + 3)² = 8² + (7x - 6)²

Expanding using the algebraic identity:

16x² + 24x + 9 = 49x² - 84x + 36

33x² - 108x + 27 = 0

3(11x² - 36x + 9) = 0

Using the quadratic formula:

x = (-b ± √(b² - 4ac)) / 2a

Substitute the values a = 11, b = -36, and c = 9:

x = (-(-36) ± √((-36)² - 4(11)(9))) / 2(11)

x = (36 ± √(1296 - 396)) / 22

x = (36 ± √900) / 22

x = (36 ± 30) / 22

x = (36 + 30) / 22 = 3

x = (36 - 30) / 22 = 3/11

x = 3 and x = 3/11.

Thus, the radius of the given circle is x = 3 and x = 3/11.

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The probability of event A occurring is 0. 65. The probability of events A and B occurring together is 0. 33. Events A and B are independent. The probability of event B is 0.68.

Probability:

The probability of an event is a number that indicates the probability of the event occurring. Expressed as a number between 0 and 1 or as a percent sign between 0% and 100%. The more likely an event is to occur, the greater its probability. The probability of an impossible event is 0; the probability of a certain event occurring is 1. The probabilities of two complementary events A and B - either A occurring or B occurring - add up to 1.

Given that:

P(A) =0.65

P(A∩B) = 0.33

And event A and B are independent.

Using the probability formula:

  P(B) = 1 - P(A) + P(A∩B)

⇒ P(B) = 1 - 0.65 + 0.33

⇒ P(B) = 0.68

Therefore , the probability of event B is 0.68.

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Answer:

y = -3/8x + 3

Step-by-step explanation:

The slope intercept form is y = mx + b

Our equation is 3x + 8y = 24

3x + 8y = 24

8y = -3x + 24

y = -3/8x + 3

So, the answer is y = -3/8x + 3

The angle of depression from the top of the center support to the end of the support cable is approximately 16.59 degrees.

The angle of depression from the top of the center support to the end of the support cable can be found using trigonometry. Let's call this angle "x". Using the given information, we can form a right triangle with the center support, the end of the support cable, and a point directly below the center support on the ground.

The height of the center support, 297 meters, is the opposite side of the right triangle, while the distance from the center of the bridge to the connection of the suspension cable, 995 meters, is the adjacent side. Using the tangent function, we can calculate the angle of depression as follows:

tan(x) = opposite/adjacent

tan(x) = 297/995

x = tan^-1(297/995)

Using a calculator, we can find that x is approximately 16.59 degrees. Therefore, the angle of depression from the top of the center support to the end of the support cable is approximately 16.59 degrees.

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A bin can hold 28 pounds. each toy car weighs 7 ounces., so the bin can hold 64 toy cars.

To determine the number of toy cars the bin can hold, we must first convert the weight limit of the bin and the weight of the toy cars to a uniform unit of measure.

We'll then divide the weight limit of the bin by the weight of one toy car. After that, we'll multiply the resulting value by the number of ounces in one pound (16).

Here's how to solve the problem:

1 pound = 16 ounces

Therefore, a bin that can hold 28 pounds can hold:28 × 16 = 448 Ounces

The weight of one toy car is 7 ounces.

Divide the weight limit of the bin (448 ounces) by the weight of one toy car (7 ounces):

448 ÷ 7 = 64

Therefore, the bin can hold 64 toy cars.

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Lisa's percentile rank is approximately 88%.

Percentile rank is a statistical measure that indicates the percentage of scores that fall below a particular score in a given distribution of data. It is commonly used to describe the relative position of a particular score in a set of scores.

If Lisa's score was 83 and that score was the 29th score from the top in a class of 240 scores, then her percentile rank can be calculated using the following formula:

Percentile Rank = [(Number of scores below Lisa's score) ÷ (Total number of scores)] × 100

Percentile Rank = [(240 - 29) ÷ 240] × 100

Percentile Rank = (211 ÷ 240) × 100

Percentile Rank = 0.8792 × 100

Percentile Rank ≈ 88 (rounded to the nearest whole number)

Therefore, her percentile rank is approximately 88%.

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Answer: 119.38

Step-by-step explanation:

C = 2πr = 2·π·19 ≈ 119.38052

Answer:

10. Other number is 12.

Explanation:

The prime factors of 120 are 2*2*2*3*5

To end up with even numbers, the odd numbers must be multiplied by even numbers. The only even numbers are 2s, while there are two odd numbers, 3 and 5.

So we MUST be talking about 2*3 and 2*5 with a 2 left over. That’s 6 and 10, which are by no stretch of the imagination consecutive. But we can either double the 10 (giving us 6 and 20, even less “consecutive”) or the 6.

Double 6 and we have 12. 10 and 12 are consecutive even numbers, because you can add two to get the next one.

There are often other methods that can converge faster than bisection method.

Advantages and disadvantages of the secant method and bisection method are:Advantages of the secant method:The secant method is faster than the bisection method since it converges faster to the root.It needs less iterations to get the root.Disadvantages of the secant method:The secant method may not converge to a root.In comparison to bisection method, it is more complex and therefore requires more computational time and effort.Advantages of the bisection method:It is more straightforward and simple to implement.In the case of continuous functions, it will always converge to a root.Disadvantages of the bisection method:It is much slower and requires more iterations to converge to a root than the secant method.There are often other methods that can converge faster than bisection method.

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Answer:

345.6

Or 345 full cubes

Step-by-step explanation:

To answer this question we first need to find the volume of the cuboid!

To find volume we use the equation...

For the cuboid we are given the dimensions 60, 24 and 30 so we just need to multiply them...

We now need to the the volume of the cube which we can just do by cubing the value given

We now need to divide the two results together to find out how many cubes would fit...

Hope this helps, have a lovely day!

Answer:

The given data suggests that there is a positive correlation - as grade level increases, so does the expected hours of homework.

Step-by-step explanation:

It is difficult to accurately predict the expected hours of homework for 5th and 8th graders based on the given data. However, we can try to make an estimate by assuming a linear correlation between grade level and hours of homework.

Using the given data, we can plot a scatter plot with grade level on the x-axis and hours of homework on the y-axis. We can then draw a line of best fit through the data points to estimate the correlation.

Assuming a linear correlation, we can use the slope-intercept form of a line to calculate the expected hours of homework for 5th and 8th graders. The equation of the line of best fit is:

y = 2.5x - 6.5

where y is the expected hours of homework per week and x is the grade level. Using this equation, we can estimate that the expected hours of homework per week for 5th graders would be:

y = 2.5(5) - 6.5 = 6 hours

And the expected hours of homework per week for 8th graders would be:

y = 2.5(8) - 6.5 = 16 hours

It is important to note that this is just an estimate based on the given data, and actual hours of homework may vary depending on the school and curriculum.

As for the correlation between grade level and hours of homework per week, the given data suggests that there is a positive correlation - as grade level increases, so does the expected hours of homework. This is likely due to the increasing complexity and workload of the curriculum as students progress through school. However, it is important for schools to ensure that the amount of homework assigned is reasonable and does not cause undue stress or negatively impact students' well-being.

Tο shift the graph tο the right, we can simply add a pοsitive cοnstant tο the x values οf each pοint befοre plοtting them. Fοr example, if we want tο shift the graph tο the right by 2 units.

The intersectiοn οf twο number lines creates a twο-dimensiοnal plane knοwn as a cοοrdinate plane. The x-axis, a hοrizοntal number line, and the y-axis, a vertical number line, are twο examples οf these number lines.

Tο graph the equatiοn 5x - 3y = 18 οn a cοοrdinate plane, we can fοllοw these steps:

1. Sοlve fοr y in terms οf x:

5x - 3y = 18

-3y = -5x + 18

y = (5/3)x - 6

2. Chοοse sοme values fοr x and use the equatiοn tο find the cοrrespοnding y values. Fοr example, we can chοοse x = 0, 3, and 6:

When x = 0: y = (5/3)(0) - 6 = -6

When x = 3: y = (5/3)(3) - 6 = -3

When x = 6: y = (5/3)(6) - 6 = 2

3. Plοt the pοints (0, -6), (3, -3), and (6, 2) οn the cοοrdinate plane.

4. Draw a straight line passing thrοugh these three pοints. This line represents the graph οf the equatiοn 5x - 3y = 18.

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Under the tested hypothesis, the outcome of the simulation has a probability of 8%. With a significance level of 1%, you should fail to reject the null hypothesis.

In statistics, the p-value is a measure of the strength of evidence against a null hypothesis. It is the probability of obtaining a test statistic as extreme or more extreme than the observed value, assuming that the null hypothesis is true.

In hypothesis testing, the null hypothesis is a statement about a population parameter that is being tested. The alternative hypothesis is another statement about the population parameter that is being tested as an alternative to the null hypothesis. The p-value is used to determine whether the observed data provides strong evidence against the null hypothesis in favor of the alternative hypothesis.

Under the tested hypothesis, the outcome of the simulation has a probability of 8%. With a significance level of 1%, you should fail to reject the null hypothesis.

The p-value is the probability of obtaining a result as extreme or more extreme than the observed result, assuming the null hypothesis is true. In this case, the null hypothesis is that the percentage of people who voted is less than or equal to 65%. Since the p-value (0.08) is greater than the significance level of 1%, we fail to reject the null hypothesis. This means that we do not have enough evidence to conclude that the percentage of people who voted is greater than 65%.

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The result of the simulation has an 8% probability under the tested hypothesis. You should be unable to deny the null hypothesis at a significance level of 1%.

The p-value in statistics is a gauge of the weight of proof opposing a null hypothesis. In the event that the null hypothesis is correct, it is the likelihood of getting a test statistic that is equally extreme or more extreme than the observed value.

The null hypothesis in hypothesis testing is a claim made about the demographic parameter under investigation. In place of the null hypothesis, a different claim is made about the demographic parameter in the alternative hypothesis. The p-value is used to assess whether the recorded data strongly support the alternative hypothesis over the null hypothesis.

The p-value, under the assumption that the null hypothesis is correct, is the likelihood of getting a result that is equally extreme or more extreme than the observed result. In this instance, the null hypothesis is that less than or equivalent to 65% of eligible voters cast ballots.

We are unable to reject the null hypothesis because the p-value (0.08), which is higher than the 1% threshold of significance, is not zero. This implies that we lack sufficient evidence to draw the conclusion that more than 65% of eligible voters cast ballots.

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For (a), 105 can be written as the product of two fractions, one of which is equal to 1. The expression would be:

105 = (105/1) × (1/1)

This expression can be simplified to:

105 = 105

For (b), the expression can be written as the product of two fractions, one of which is equal to 1. The expression would be:

5/2 = (5/2) × (1/1)

This expression can be simplified to:

5/2

For (c), the expression can be written as the product of two fractions, one of which is equal to 1. The expression would be:

18 = (18/1) × (1/1)

This expression can be simplified to:

18 = 18

[tex]{\Large \begin{array}{llll} y=x^2+ax+b \end{array}} \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} x=0\\ y=-7 \end{cases}\implies -7=0^2+a(0)+b\implies \boxed{-7=b} \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} x=7\\ y=0 \end{cases}\implies 0=7^2+a7+\stackrel{b}{(-7)}\implies 0=49+7a-7 \\\\\\ 0=42+7a\implies -42=7a\implies \cfrac{-42}{7}=a\implies \boxed{-6=a} \\\\\\ ~\hfill {\Large \begin{array}{llll} y=x^2-6x-7 \end{array}}~\hfill[/tex]

now, let's get the "vertex" using the coefficients.

[tex]\textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{1}x^2\stackrel{\stackrel{b}{\downarrow }}{-6}x\stackrel{\stackrel{c}{\downarrow }}{-7} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ -6}{2(1)}~~~~ ,~~~~ -7-\cfrac{ (-6)^2}{4(1)}\right) \implies \left( - \cfrac{ -6 }{ 2 }~~,~~-7 - \cfrac{ 36 }{ 4 } \right) \\\\\\ \left( 3 ~~~~ ,~~~~ -7 -9 \right)\implies {\Large \begin{array}{llll} (3~~,~-16) \end{array}}[/tex]