Which data set is the most clustered around its mean?
A.4, 10, 8, 6
B.11, 3, 10, 4
C.2, 9 ,13, 4
D.11, 2, 9, 6

Answers

Answer 1

The data set is the most clustered around its mean is

A.4, 10, 8, 6

How to find the data

To determine which data set is the most clustered around its mean, we can calculate the mean and the standard deviation for each data set.

calculate the mean and standard deviation for each data set

Data Set A: 4, 10, 8, 6

Mean: (4 + 10 + 8 + 6) / 4 = 28 / 4 = 7

Standard Deviation: √[(4-7)^2 + (10-7)^2 + (8-7)^2 + (6-7)^2] / 4 ≈ 1.58

Data Set B: 11, 3, 10, 4

Mean: (11 + 3 + 10 + 4) / 4 = 28 / 4 = 7

Standard Deviation: √[(11-7)^2 + (3-7)^2 + (10-7)^2 + (4-7)^2] / 4 ≈ 3.87

Data Set C: 2, 9, 13, 4

Mean: (2 + 9 + 13 + 4) / 4 = 28 / 4 = 7

Standard Deviation: √[(2-7)^2 + (9-7)^2 + (13-7)^2 + (4-7)^2] / 4 ≈ 4.27

Data Set D: 11, 2, 9, 6

Mean: (11 + 2 + 9 + 6) / 4 = 28 / 4 = 7

Standard Deviation: √[(11-7)^2 + (2-7)^2 + (9-7)^2 + (6-7)^2] / 4 ≈ 3.27

Comparing the standard deviations, we find that data set A has the smallest standard deviation of approximately 1.58. this indicates that the data points in Data Set A are the closest to the mean, making it the most clustered around its mean.

Therefore, the answer is A. 4, 10, 8, 6.

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Related Questions



2. Describe a rigid motion or composition of rigid motions that maps the rectangular bench at (0, 10)and

the adjacent flagpole onto the other short rectangular bench and flagpole.

Answers

Answer:

See Explanation

Step-by-step explanation:

Given

Let the bench be B and the flagpole be T.

So:

[tex]B = (0,10)[/tex] --- given

The flagpole is represented by the triangular shape labelled T.

So, we have:

[tex]T = (6,9)[/tex]

See attachment for the rectangular bench and the flagpole

From the attached image, the location of the other bench is:

[tex]B' = (0,-10)[/tex]

And the location of the other flagpole is:

[tex]T' = (-6,9)[/tex]

So, we have:

[tex]B = (0,10)[/tex] ==> [tex]B' = (0,-10)[/tex]

[tex]T = (6,9)[/tex] ==> [tex]T' = (-6,9)[/tex]

When a point is reflected from [tex](x,y)[/tex] to [tex](x,-y)[/tex], the transformation rule is reflection across x-axis.

So the rigid transformation that takes [tex]B = (0,10)[/tex] to [tex]B' = (0,-10)[/tex] is: reflection across x-axis.

When a point is reflected from [tex](x,y)[/tex] to [tex](-x,y)[/tex], the transformation rule is reflection across y-axis.

So the rigid transformation that takes [tex]T = (6,9)[/tex] to [tex]T' = (-6,9)[/tex] is: reflection across y-axis.

120, 60, 30, 15, ...
8th term:
10th term:

Answers

Answer:

8th term would be .9375

10th term would be .234375

Step-by-step explanation:

YO PLEASE HELP, AREA OF A TRAPEZOID

Answers

the answer is 54!!!!!

how tall is the building in the picture shown​

Answers

Answer: just figure out how many 8s are in 36 then multiply that by 66 :):)

Step-by-step explanation:

A cylindrical gas tank holds approximately 401.92 cubic feet of fuel. The radius of the tank is 4 feet. Find the height of the tank in feet. Use 3.14 for π. PLEASE HURRY I NEED THIS

Answers

Answer:

8 ft

Step-by-step explanation:

Given:

Volume = 401.92

radius = 4 feet

Volume of cylinder: [tex]\pi[/tex]r^2h

3.14 * 4^2 * h = 401.92 cubic feet

3.14 * 16 * h = 401.92 cubic feet

50.24 * h = 401.92

h = 8 ft

Angela bought a dress that cost $22.50. If the sales tax is 6.5%, how much tax did she pay? I need help with this one

Answers

your answer will be $23.96

Flordia yum, an ice cream company would like to test the hypothesis that the variance for the number of ounces of ice cream that Americans consume per month is greater than 15. A random sample of 23 people was found to have a standard deviation of 4.4 ounces for the quantity of ice cream consumed per month. The ice cream company would like to test the hypothesis using alpha= 0.05.

Answers

If the test statistic exceeds the critical value, we reject the null hypothesis and conclude that there is evidence to support the claim that the variance is greater than 15. Otherwise, if the calculated chi-square value is less than or equal to the critical chi-square value, we fail to reject the null hypothesis.

To test the hypothesis that the variance for the number of ounces of ice cream consumed per month is greater than 15, we can use a chi-square test for variance.

The null and alternative hypotheses for this test are as follows:

Null Hypothesis (H0): The variance for the number of ounces of ice cream consumed per month is equal to or less than 15.

Alternative Hypothesis (H1): The variance for the number of ounces of ice cream consumed per month is greater than 15.

Given a random sample of 23 people with a standard deviation of 4.4 ounces, we can calculate the sample variance as [tex]s^2 = (4.4)^2[/tex] = 19.36.

To perform the chi-square test, we calculate the test statistic as follows:

chi-square = (n - 1) × [tex]s^2[/tex] / [tex]\sigma^2[/tex]

where n is the sample size, [tex]s^2[/tex] is the sample variance, and [tex]\sigma^2[/tex] is the hypothesized population variance (15 in this case).

Substituting the values, we get:

chi-square = (23 - 1) × 19.36 / 15 = 29.44

Next, we compare this test statistic to the critical value from the chi-square distribution table with (n - 1) degrees of freedom (22 degrees of freedom in this case) at the given significance level (alpha = 0.05).

If the test statistic is greater than the critical value, we reject the null hypothesis and conclude that the variance for the number of ounces of ice cream consumed per month is greater than 15.

Otherwise, if the test statistic is less than or equal to the critical value, we fail to reject the null hypothesis.

Performing the chi-square test with the given values, we compare the test statistic (29.44) to the critical value from the chi-square distribution table.

If the test statistic exceeds the critical value, we reject the null hypothesis and conclude that the variance for the number of ounces of ice cream consumed per month is greater than 15. Otherwise, we fail to reject the null hypothesis.

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How many sqftwks are used when producing a K flat of cucumbers
for grafting that spent 6 weeks on a bench in the greenhouse?

Answers

The number of square feet-weeks used when producing a K flat of cucumbers for grafting is not provided.

To determine the number of sqftwks (square feet-weeks) used when producing a K flat of cucumbers for grafting that spent 6 weeks on a bench in the greenhouse, we need additional information.

The term "sqftwks" represents the product of the area in square feet and the duration in weeks. However, the specific values for the area and the duration of cucumber growth are not provided in the question.

To calculate the number of sqftwks, we would need to know the area occupied by the K flat of cucumbers and the time spent in the greenhouse. Without these specific values, it is not possible to determine the number of sqftwks used for cucumber production.

Therefore, based on the given information, we cannot calculate the number of sqftwks used when producing a K flat of cucumbers for grafting that spent 6 weeks on a bench in the greenhouse.

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Set up an integral that represents the length of the part of the parametric curve shown in the graph.
x = 9t^2 − 3t^3, y = 3t^2 − 6t
The x y-coordinate plane is given. The curve starts at the point (12, 9), goes down and left becoming more steep, changes direction at approximately the origin, goes down and right becoming less steep, changes direction at the point (6, −3), goes up and right becoming more steep, changes direction at the approximate point (12, 0), goes up and left becoming less steep, and stops at the point (0, 9).

Answers

To find the length of the parametric curve described by the equations x = 9t^2 − 3t^3 and y = 3t^2 − 6t, we can set up an integral using the arc length formula. The curve starts at point (12, 9) and ends at point (0, 9), with several changes in direction along the way.

The length of a curve can be calculated using the arc length formula. For a parametric curve defined by x = f(t) and y = g(t), the arc length can be expressed as:

L = ∫[a,b] √[(dx/dt)^2 + (dy/dt)^2] dt.

In this case, we have x = 9t^2 − 3t^3 and y = 3t^2 − 6t. To find the length of the curve, we need to determine the interval [a, b] over which t varies.

From the given information, we can see that the curve starts at (12, 9) and ends at (0, 9). By solving the equation x = 9t^2 − 3t^3 for t, we find that t = 0 at x = 0, and t = 2 at x = 12. Therefore, the interval of integration is [0, 2].

To set up the integral, we calculate the derivatives dx/dt and dy/dt, and then substitute them into the arc length formula. Simplifying the expression inside the square root and integrating over the interval [0, 2], we can evaluate the integral to find the length of the curve.

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Express 5: 8 in the ratio n : 1

Answers

Answer:

0.625 : 1

Step-by-step explanation:

5 : 8

→ Divide both sides by 8 to get it to 1

(5 ÷ 8) : (8 ÷ 8)

→ Simplify

0.625 : 1

The mean salary at a local industrial plant is $28,600$⁢28,600 with a standard deviation of $4400$⁢4400. The median salary is $25,300$⁢25,300 and the 61st percentile is $29,000$⁢29,000.

Step 2 of 5:

Based on the given information, determine if the following statement is true or false.

Joe's salary of $35,060$⁢35,060 is 1.401.40 standard deviations above the mean.

Answers

The statement Joe's salary of $35,060$⁢35,060 is 1.401.40 standard deviations above the mean is false.

To determine if the statement is true or false, we need to calculate the number of standard deviations Joe's salary of $35,060 is above the mean.

Given:

Mean salary = $28,600

Standard deviation = $4,400

Joe's salary = $35,060

To calculate the number of standard deviations above the mean, we can use the formula:

Number of standard deviations = (X - μ) / σ

Where:

X is the value we want to compare (Joe's salary)

μ is the mean

σ is the standard deviation

Plugging in the values, we have:

Number of standard deviations = (35,060 - 28,600) / 4,400

= 6,460 / 4,400

≈ 1.4727

Therefore, Joe's salary of $35,060 is approximately 1.4727 standard deviations above the mean, not 1.40.

The statement is false.

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Due soon !!!!!!!...

Answers

Answer:

50m^2

Step-by-step explanation:

A pyramid with a rectangular base has a volume of 60 cubic feet and a height of 6 feet. The width of the rectangular base is 4 feet. Find the length of the rectangular base.

Answers

Answer:

length=2.5feet

Step-by-step explanation:

Volume =Base area * height

6/60=(4*length) * 6/6

10=4length

length=10÷4

=2.5feet

The length of the pyramid is 7.5 feet.

What is a pyramid with rectangular base?

'A rectangular pyramid is a type of pyramid with the base shaped like a rectangle but the sides are shaped like a triangle. A pyramid usually has triangular sides but with different bases.'

According to the given problem,

Volume = 60 cubic feet

Height = 6 feet

Width of rectangular base = 4 feet

Let the length be x,

Volume =[tex]\frac{L * W * H}{3}[/tex]

⇒ 60 = [tex]\frac{x*6*4}{3}[/tex]

⇒ 180 = 24x

⇒ x = [tex]\frac{180}{24}[/tex]

⇒ x =[tex]\frac{45}{6}[/tex] = 7.5

Hence, we can conclude that the value of the length of the rectangular base is 7.5 feet.

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A group of students were asked if they like Math class and if they like English class.
Partial results are shown in the table.
Like Math
Don't Like Math
Like English
х
9
Don't Like English
26
у
.
Of the students who like math, 40% don't like English.
Of the students who don't like math, 75% don't like English
What is the value of x + y?
A. 35
B. 57
C. 65
D. 66

Answers

a if that’s wrong i’m vv sorry
Yea I’m pretty sure the answer A

6,336 ft. = in league

Answers

Answer:

6336 feet = 0.3476 nautical leagues

pls help will mark brainlest

Answers

Answer:

(5,7) , (2, -3) , (2,4)

Step-by-step explanation:

your first number, the x, is the horizontal number. The second, y, is the vertical number. So if you look at A, the x is 5, and you then have to go up to 7 to reach the intercept

i need help please i'll give brainly​

Answers

Answer:

60 +24 = 84

Step-by-step explanation:jjjjjhjhjhj

6 x 4 +5 x 12

A circular opening of an ice cream cone has a diameter of 23 centimeters. The height of the cone is 17 centimeters, what is the

volume of the cone?

round to hundredth decimals if necessary.

Answers

Answer:[tex]2355.31\ cm^3[/tex]

Step-by-step explanation:

Given

diameter of the cone is  [tex]d=23\ cm\quad [r=\frac{d}{2}=11.5\ cm][/tex]

The height of the cone is  [tex]h=17\ cm[/tex]

The volume of a cone is  [tex]V=\dfrac{1}{3}\pi r^2h[/tex]

Insert the values

[tex]\Rightarrow V=\dfrac{1}{3}\times \dfrac{22}{7}\times 11.5^2\times 17=2355.309\\\\\Rightarrow V\approx 2355.31\ cm^3[/tex]

Thus, the volume of the cone is [tex]2355.31\ cm^3[/tex]

If one of my angles is 130 degrees, what is the measure of the blue angle that is supplementary to it?

Answers

Step-by-step explanation:

[tex]180 - 130 = 50[/tex]

In myrtle beach, the maximum height off the water a parasailor can be is 500 feet. I the rope is 800 feet long, about how many feet from the boat is the parasailor to the nearest tenth?

Answers

Answer:

624.5 feet

Step-by-step explanation:

Calculation to determine how many feet from the boat is the parasailor

Based on the information given we would make use of Pythagorean theorem to determine how many feet from the boat is the parasailor using this formula

a²+b²=c²

First step is to plug in the formula by substituting the given value

500²+b²=800²

Second step is to evaluate the exponent

250,000+b²=640,000

Third step is to substract 250,000 from both side and simplify

250,000+b²-250,000=640,000-250,000

b²=390,000

Now let determine how many feet from the boat is the parasailor

Parasailor feet=√b²

Parasailor feet=√390,000

Parasailor feet=b=624.49

Parasailor feet=b=624.5 feet (Approximately)

Therefore how many feet from the boat is the parasailor will be 624.5 feet

does someone know the answer?

Answers

Answer:

I think it is the last answer.

Step-by-step explanation:

There can only be one vertex.




122. Density Function of Y Let X = N(0, 1). Calculate the density function of Y = √|X|.

Answers

The density function of Y, denoted as [tex]f_Y[/tex](y), is given by [tex]f_Y[/tex](y) = 2y × φ(y²), where φ(y) is the standard normal density function.

To find the density function of Y = √|X|, we need to calculate the cumulative distribution function (CDF) of Y and then differentiate it to obtain the density function.

Start with the CDF of Y:

[tex]f_Y[/tex](y) = P(Y ≤ y) = P(√|X| ≤ y)

Since X is normally distributed with mean 0 and variance 1, we know that X follows a standard normal distribution.

We can rewrite the CDF of Y in terms of X:

[tex]f_Y[/tex](y) = P(-y ≤ √|X| ≤ y) = P(X ≤ y²) - P(X ≤ -y²)

Using the standard normal distribution table, we can find the probabilities:

[tex]f_Y[/tex](y) = Φ(y²) - Φ(-y²)

Differentiating the CDF with respect to y, we obtain the density function of Y:

[tex]f_Y[/tex](y) = d/dy[[tex]f_Y[/tex](y)] = d/dy[Φ(y²) - Φ(-y²)]

Simplifying the derivative, we get:

[tex]f_Y[/tex](y) = 2y × φ(y²)

where φ(y) is the standard normal density function.

Thus, the density function of Y is [tex]f_Y[/tex](y) = 2y × φ(y²), where φ(y) is the standard normal density function.

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select the expression that is equivalent to: 3 square root 1089

Answers

By finding the square root of 1089, we determine that it is equal to 33. Multiplying 3 by 33 results in 99, which is the equivalent expression to 3√1089.

To simplify the expression, we need to find the square root of 1089. The square root of 1089 is 33 because 33 * 33 = 1089.

Now, multiplying 3 by √1089 gives us 3 * 33, which equals 99. Therefore, the expression 3√1089 is equivalent to 99.

When we have a cube root (∛) in the original expression, we need to find the number that, when multiplied by itself three times, equals the given value. However, in this case, we have a square root (√) in the expression, which means we need to find the number that, when multiplied by itself once, equals the given value.

By finding the square root of 1089, we determine that it is equal to 33. Multiplying 3 by 33 results in 99, which is the equivalent expression to 3√1089.

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A decision problem X is self-solvable if X = L(MX) for some polynomial-time oracle TM M, whose oracle queries are always strictly shorter than its input. In other words, when M is executed on an input of length n, it queries its oracle only on strings of length less than n. This is a strange situation, where M has oracle access to the problem that it is trying to solve. But when M is trying to determine whether x € X, it cannot simply query its oracle on x for the answer. ከ. (a) Show that TQBF is self-solvable. Be explicit about what assumptions are you making about how for- mulas are encoded into bit strings. (b) Prove that if L is self-solvable then L E PSPACE.

Answers

(a) The TQBF oracle used by M satisfies the condition for self-solvability.

It can handle formulas of length strictly shorter than the input length, ensuring that M's oracle queries are always strictly shorter than its input.

To show that TQBF (True Quantified Boolean Formula) is self-solvable, we need to demonstrate that there exists a polynomial-time oracle Turing machine (TM) M that can solve TQBF using an oracle for TQBF.

Assuming that formulas in TQBF are encoded into bit strings in a standard way, we can construct the TM M as follows:

On input x (the encoded TQBF formula), M starts by generating all possible truth assignments for the variables in the formula.

For each truth assignment, M constructs the corresponding quantified Boolean formula and queries the TQBF oracle with this formula.

If the oracle returns "true" for any truth assignment, M accepts x. Otherwise, if the oracle returns "false" for all truth assignments, M rejects x.

Hence, TQBF (True Quantified Boolean Formula) is self-solvable.

(b) If a language L is self-solvable, it implies that L is in PSPACE (polynomial space complexity class).

To prove that if L is self-solvable, then L is in PSPACE, we can show that a polynomial-time oracle TM M with oracle access to L can be simulated by a polynomial-space Turing machine.

Let M' be a polynomial-space Turing machine that simulates M with oracle access to L. Since L is self-solvable, M' can query the oracle on inputs shorter than its own input.

We can construct a polynomial-space Turing machine M'' that simulates M' without the need for an oracle. M'' uses its own polynomial space to simulate the computation of M'. Whenever M' queries the oracle on an input, M'' runs its own simulation for that input length using its available space.

Since M'' only requires polynomial space and simulates the behavior of M', it follows that L is in PSPACE.

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The figure below is a net for a triangular pyramid.
If all the triangles are equilateral, what is the surface area of
the pyramid, in square centimeters?

Answers

Answer:

339.36 cm^2

Step-by-step explanation:

14*12.12=169.68

169.68/2=84.84

84.84= the area of one triangle

(84.84)4=339.36

I'm pretty sure, but check my math if it's wrong

can someone plz help me plz plz

Answers

Answer:

56.52

Step-by-step explanation:

C = 2 x pi x r

r = 9

C = 2 x pi x r

multiply 2 by 9

C = 18 x pi

C = 56. 52

Please show me step by step how to do this

Answers

Answer:

The next three terms of the sequence are 17, 21 and 25.

The 300th term of the sequence is 1197.

Step-by-step explanation:

The statement describes an arithmetic progression, which is defined by following formula:

[tex]p(n) = p_{1}+r\cdot (n-1)[/tex] (1)

Where:

[tex]p_{1}[/tex] - First element of the sequence.

[tex]r[/tex] - Progression rate.

[tex]n[/tex] - Index of the n-th element of the sequence.

[tex]p(n)[/tex] - n-th element of the series.

If we know that [tex]p_{1} = 1[/tex], [tex]n = 2[/tex] and [tex]p(n) = 5[/tex], then the progression rate is:

[tex]r = \frac{p(n)-p_{1}}{n-1}[/tex]

[tex]r = 4[/tex]

The set of elements of the series are described by [tex]p(n) = 1 + 4\cdot (n-1)[/tex].

Lastly, if we know that [tex]n = 300[/tex], then the 300th term of the sequence is:

[tex]p(n) = 1 + 4\cdot (n-1)[/tex]

[tex]p(n) = 1197[/tex]

And the next three terms of the sequence are:

n = 5

[tex]p(n) = 1 + 4\cdot (n-1)[/tex]

[tex]p(n) = 17[/tex]

n = 6

[tex]p(n) = 1 + 4\cdot (n-1)[/tex]

[tex]p(n) = 21[/tex]

n = 7

[tex]p(n) = 1 + 4\cdot (n-1)[/tex]

[tex]p(n) = 25[/tex]

The next three terms of the sequence are 17, 21 and 25.

The 300th term of the sequence is 1197.

Solve the following system of equations using Desmos.....x-3y=-2 and x+3y=16

Answers

Answer:

(7, 3 )

Step-by-step explanation:

Given the 2 equations

x - 3y = - 2 → (1)

x + 3y = 16 → (2)

Adding (1) and (2) term by term will eliminate the y- term

2x + 0 = 14

2x = 14 ( divide both sides by 2 )

x = 7

Substitute x = 7 into either of the 2 equations and solve for y

Substituting into (2)

7 + 3y = 16 ( subtract 7 from both sides )

3y = 9 ( divide both sides by 3 )

y = 3

solution is (7, 3 )

Sam drinks water at an incredible rate. She drinks 2 1/5 liters of water every 3/4 of an hour. Sam drinks water at a constant rate. About how much water does Sam drink per hour?

Answers

Answer:

2.9 litres

or

2 14/15 litres

Step-by-step explanation:

To determine the amount of water Sam drinks in an hour, divide the total amount drunk by the number of hours

Amount drank in an hour = total litres drunk / time

[tex]2\frac{1}{5}[/tex] ÷ [tex]\frac{3}{4}[/tex]

[tex]\frac{11}{5}[/tex] × [tex]\frac{4}{3}[/tex] = [tex]\frac{44}{15}[/tex] = 2[tex]\frac{14}{15}[/tex] litres

Let X is uniformly distributed over (0,1) and Y is exponentially distributed with parameter lambda = 2. Furthermore assume X and Y are independent. The cumulative distribution of Z = X + Y is P{Z lessthanorequalto a} = P{X + Y lessthanorequalto a} =___________________________for 0 < a < 1 P{Z lessthanorequalto a} = P{X + Y lessthanorequalto a} =___________________________for 0 < a < infinity The cumulative distribution of T = x/y is P({T lessthanorequalto a} = P{X/a lessthanorequalto Y} =___________________________for_________< a

Answers

To find the cumulative distribution function (CDF) of Z = X + Y, where X is uniformly distributed over (0,1) and Y is exponentially distributed with parameter lambda = 2, we can use the properties of independent random variables.

For 0 < a < 1, we have:

P(Z ≤ a) = P(X + Y ≤ a)

Since X and Y are independent, we can write this as:

P(Z ≤ a) = ∫∫ P(X ≤ x, Y ≤ a - x) dxdy

Since X is uniformly distributed over (0,1) and Y is exponentially distributed with parameter lambda = 2, we have their respective probability density functions (PDFs):

fX(x) = 1, 0 ≤ x ≤ 1

fY(y) = 2e^(-2y), y ≥ 0

Now, we can calculate the CDF of Z:

P(Z ≤ a) = ∫∫ P(X ≤ x, Y ≤ a - x) dxdy

= ∫∫ fX(x) * fY(y) dxdy, since X and Y are independent

= ∫∫ 1 *[tex]2e^(-2y)[/tex] dxdy, for 0 ≤ x ≤ 1 and y ≥ 0

Integrating with respect to x from 0 to 1 and with respect to y from 0 to a - x, we get:

P(Z ≤ a) = ∫[0,1]∫[0,a-x] 1 * 2[tex]e^(-2y)[/tex]dydx

= ∫[0,1] [[tex]-e^(-2y)[/tex]] [0,a-x] dx

= ∫[0,1] (1 - [tex]e^(-2(a-x)[/tex])) dx

Evaluating the integral, we have:

P(Z ≤ a) = [x - [tex]xe^(-2(a-x))[/tex]] [0,1]

= (1 - e^(-2a))

Therefore, the cumulative distribution function (CDF) of Z is:

P(Z ≤ a) = [tex](1 - e^(-2a)),[/tex] for 0 < a < 1

For 0 < a < ∞, the cumulative distribution function of Z remains the same:

P(Z ≤ a) = (1 - e^(-2a)), for 0 < a < ∞

Now, let's move on to the cumulative distribution function of T = X/Y.

P(T ≤ a) = P(X/Y ≤ a)

Since X and Y are independent, we can write this as:

P(T ≤ a) = ∫∫ P(X/y ≤ a) fX(x) * fY(y) dxdy

Using the given information that X is uniformly distributed over (0,1) and Y is exponentially distributed with parameter lambda = 2, we can substitute their respective PDFs:

P(T ≤ a) = ∫∫ P(X/y ≤ a) * 1 * [tex]2e^(-2y)[/tex]dxdy

= ∫∫ P(X ≤ ay) * 1 * [tex]2e^(-2y)[/tex]dxdy

Now, we need to determine the range of integration for x and y. Since X is between 0 and 1, and Y is greater than or equal to 0, we have:

0 ≤ x ≤ 1

0 ≤ y

Using these limits, we can calculate the CDF of T:

P(T ≤ a) = ∫[0,1]∫[0,∞] P(X ≤ ay) * 1 * [tex]2e^(-2y)[/tex] dydx

To evaluate this integral, we need to consider the range of values for ay. Since a can be any positive real number, ay can range from 0 to ∞.

P(T ≤ a) = ∫[0,1]∫[0,∞] P(X ≤ ay) * 1 * 2[tex]e^(-2y)[/tex] dydx

= ∫[0,1]∫[0,∞] (ay) * 1 * 2[tex]e^(-2y)[/tex] dydx, for ay ≥ 0

Integrating with respect to y from 0 to ∞ and with respect to x from 0 to 1, we have:

P(T ≤ a) = ∫[0,1]∫[0,∞] (ay) * 1 * 2[tex]e^(-2y)[/tex]dydx

= ∫[0,1] (2a / (4 + a^2)) dx

Evaluating the integral, we get:

P(T ≤ a) = (2a / (4 + [tex]a^2)),[/tex] for a > 0

Therefore, the cumulative distribution function (CDF) of T is:

P(T ≤ a) = (2a / (4 + [tex]a^2)),[/tex] for a > 0

Learn more about cumulative distribution function here:

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