2. Find the Area of the triangle.
as do
12 yd
15 yd

Final answer:

The events A and B are mutually exclusive. The events A and C are not mutually exclusive. Two events A and B are independent events.

Explanation:

The events A and B are mutually exclusive because they cannot happen at the same time. P(A AND B) = 0.

No, A and C are not mutually exclusive because they can occur at the same time. In fact, C includes all of the outcomes of A; if the card chosen is blue it is also (red or blue). P(A AND C) = P(A) = 20.

Two events A and B are independent events if the knowledge that one occurred does not affect the chance the other occurs. For example, the outcomes of two roles of a fair die are independent events. The outcome of the first roll does not change the probability for the outcome of the second roll. To show two events are independent, you must show only one of the above conditions. If two events are not independent, then we say that they are dependent events. Sampling may be done with replacement or without replacement. With replacement: If each member of a population is replaced after it is picked, then that member has the possibility of being chosen more than once. When sampling is done with replacement, then events are considered to be independent, meaning the result of the first pick will not change the probabilities for the second pick.

Answer:

it d (42)

Step-by-step explanation:

Answer:

d: 42

Step-by-step explanation:

Answer:

a) [tex]\vec v_{B} - \vec v_{C} = 3\, i - 17\,j[/tex], b) [tex]\theta = 79.992^{\textdegree}[/tex] (clockwise)

Step-by-step explanation:

The resultant velocity of the boat must be:

[tex]\vec v_{B} = 15\,i[/tex]

Likewise, the velocity of the current is:

[tex]\vec v_{C} = 2\,i + 17\,j[/tex]

a) The intended velocity of the boat is:

[tex]\vec v_{B} - \vec v_{C} = (15-2)\,i + (0-17)\,j[/tex]

[tex]\vec v_{B} - \vec v_{C} = 3\, i - 17\,j[/tex]

b) The direction of the boat is:

[tex]\theta = \tan^{-1}\left(\frac{17}{3} \right)[/tex] (clockwise)

[tex]\theta = 79.992^{\textdegree}[/tex] (clockwise)

answer:

1.94117647058

Step-by-step explanation:

33/50 her free throw % is 66% at the moment based on how many she missed you'd do 33 / by 17 you get 1.94117647058

Answer:

Step-by-step explanation:

Find the volume of a sphere 12 yards a crossed

Answer:

99% confidence interval for the difference between the mean thicknesses produced by the two methods is [0.099 μm , 0.901 μm].

Step-by-step explanation:

We are given that in a study comparing various methods of gold plating, 7 printed circuit edge connectors were gold-plated with control-immersion tip plating. The average gold thickness was 1.5 μm, with a standard deviation of 0.25 μm.

Five connectors were masked and then plated with total immersion plating. The average gold thickness was 1.0 μm, with a standard deviation of 0.15 μm.

Firstly, the pivotal quantity for 99% confidence interval for the difference between the population mean is given by;

                              P.Q. = [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex]  ~ [tex]t__n__1+_n__2-2[/tex]

where, [tex]\bar X_1[/tex] = average gold thickness of control-immersion tip plating = 1.5 μm

[tex]\bar X_2[/tex] = average gold thickness of total immersion plating = 1.0 μm

[tex]s_1[/tex] = sample standard deviation of control-immersion tip plating = 0.25 μm

[tex]s_2[/tex] = sample standard deviation of total immersion plating = 0.15 μm

[tex]n_1[/tex] = sample of printed circuit edge connectors plated with control-immersion tip plating = 7

[tex]n_2[/tex] = sample of connectors plated with total immersion plating = 5

Also, [tex]s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }[/tex]   =  [tex]\sqrt{\frac{(7-1)\times 0.25^{2}+(5-1)\times 0.15^{2} }{7+5-2} }[/tex]  = 0.216

Here for constructing 99% confidence interval we have used Two-sample t test statistics as we don't know about population standard deviations.

So, 99% confidence interval for the difference between the mean population mean, ([tex]\mu_1-\mu_2[/tex]) is ;

P(-3.169 < [tex]t_1_0[/tex] < 3.169) = 0.99  {As the critical value of t at 10 degree of

                                              freedom are -3.169 & 3.169 with P = 0.5%}  

P(-3.169 < [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] < 3.169) = 0.99

P( [tex]-3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] < [tex]{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}[/tex] < [tex]3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ) = 0.99

P( [tex](\bar X_1-\bar X_2)-3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] < ([tex]\mu_1-\mu_2[/tex]) < [tex](\bar X_1-\bar X_2)+3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ) = 0.99

99% confidence interval for ([tex]\mu_1-\mu_2[/tex]) =

[ [tex](\bar X_1-\bar X_2)-3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] , [tex](\bar X_1-\bar X_2)+3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ]

= [ [tex](1.5-1.0)-3.169 \times {0.216\sqrt{\frac{1}{7}+\frac{1}{5} } }[/tex] , [tex](1.5-1.0)+3.169 \times {0.216\sqrt{\frac{1}{7}+\frac{1}{5} } }[/tex] ]

= [0.099 μm , 0.901 μm]

Therefore, 99% confidence interval for the difference between the mean thicknesses produced by the two methods is [0.099 μm , 0.901 μm].

Answer:

The 99% of confidence intervals for difference between the mean thicknesses produced by the two methods.

( 0.17971  , 0.82028)

Step-by-step explanation:

Step:-(i)

Given data the average gold thickness was 1.5 μm, with a standard deviation of 0.25 μ m.

Given sample size n₁ = 7

mean of first sample x₁⁻ =1.5 μ m.

Standard deviation of first sample S₁ = 0.25 μ m

Given data  Five connectors were masked and then plated with total immersion plating. The average gold thickness was 1.0 μ m, with a standard deviation of 0.15 μ m.

Given second sample size n₂ = 5

The mean of second sample x⁻₂ =  1.0 μ m

Standard deviation of first sample S₂ = 0.15 μ m

Level of significance ∝ =0.01 or 99%

Degrees of freedom γ = n₁+ n₂ -2 = 7+5-2=10

tabulated value t = 2.764

Step(ii):-

The 99% of confidence intervals for μ₁- μ₂ is determined by

(x₁⁻ - x⁻₂)  - z₀.₉₉ Se((x₁⁻ - x⁻₂) ,  (x₁⁻ - x⁻₂)  + z₀.₉₉ Se((x₁⁻ - x⁻₂)

where         [tex]se(x^{-} _{1}-x^{-} _{2} ) = \sqrt{\frac{s^2_{1} }{n_{1} } +\frac{s^2_{2} }{n_{2} } }[/tex]

                   [tex]se(x^{-} _{1}-x^{-} _{2} ) = \sqrt{\frac{0.25^2 }{7 } +\frac{0.15^2 }{5 } } = 0.115879[/tex]

[1.5-1.0 -  2.764 (0.115879) , (1.5+1.0) + 2.764(0.115879 ]

(0.5-0.32029,0.5+0.32029

( 0.17971  , 0.82028)

Conclusion:-

The 99% of confidence intervals for μ₁- μ₂ is determined by

( 0.17971  , 0.82028)

Answer:

8π yards

Step-by-step explanation:

A circle subtends a total angle of 360 ° from its center.The length of an arc is directly proportional to the angle it subtends from the circle's center. The arc's length can therefore be calculated as:

C=πd=2πr

Where C is circumference, d is diameter and r is radius. Given r as 16 yards then the arc length which is equivalent to circumference is given as

[tex]C=\frac {90}{360}*2*\pi*16=8\pi[/tex]

Answer:

2,144.7 i just answered the other one is wrong

Step-by-step explanation:

Answer: C. Because the graphs of the two equations overlap each other.

Step-by-step explanation: I took the test its right! Hope this helps

Step-by-step explanation:

[tex](7m + 8)(4m + 1) \\ = 7m(4m + 1) + 8(4m + 1) \\ = 28 {m}^{2} + 7m + 32m + 8 \\ \purple { \bold{= 28 {m}^{2} + 39m + 8}}[/tex]

Answer:

You can use foil or the box method.

Step-by-step explanation:

Sorry it looks so bad :P

Answer:

a) The mean of the sampling distribution of p is 0.4.

The standard deviation of the sampling distribution of p is 0.0538.

b) The shape of the sampling distribution of p is approximately normal because the sample size is large.

c) z=0.19

Step-by-step explanation:

We have a random sample of size n=83, drawn from a binomial population with proabiliity p=0.4. We have to compute the characteristic of the sampling distribution of the sample proportion.

a) The mean of the sampling distribution is equal to the mean of the distribution p:

[tex]\mu_p=p=0.4[/tex]

The standard deviation of the sampling distribution is:

[tex]\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.4*0.6}{83}}=\sqrt{0.0029}=0.0538[/tex]

b) The shape of sampling distribution with enough sample size tend to be approximately normal. In this case, n=83 is big enough for a binomial distribution.

c) The z-score fof p-0.41 can be calculated as:

[tex]z=\dfrac{p-\mu_p}{\sigma_p}=\dfrac{0.41-0.4}{0.0538}=\dfrac{0.01}{0.0538}=0.19[/tex]

The mean of the sampling distribution is 0.4 and the standard deviation is 0.0538.

From the information given, the sample is drawn from a binomial population with probability of success 0.4. Therefore, the mean is 0.4.

The standard deviation will be:

= [(✓0.4 × ✓0.6) / ✓83)]

= 0.0538

Furthermore, the shape of the sampling distribution of p is approximately normal because the sample size is large.

The standard normal z-score will be:

= (0.41 - 0.4) / 0.0538

= 0.10/0.0538

= 0.19.

Therefore, the standard normal z-score is 0.19.

Learn more about sampling on:

https://brainly.com/question/24466382

Answer:

2

Step-by-step explanation:

Answer: 9?

Step-by-step explanation: I think.

Answer:

He has 5 children

Step-by-step explanation:One boy is a brother to all of them

A number is an arithmetic value that is used to represent a quantity and calculate it. The number of children that Mr. Smith will have is 5.

A number is an arithmetic value that is used to represent a quantity and calculate it. Numericals are written symbols that represent numbers, such as "3."

As it is mentioned that Mr. Smith had 4 daughters and each of the daughters had a brother. Now, if Mr. Smith had just a single son then he will be a brother to every daughter. Therefore, the number of children that Mr. Smith will have is 5.

Learn more about Numbers:

https://brainly.com/question/17429689

Answer:

We conclude that the calibration point is set too low.

Step-by-step explanation:

We are given that NIST technicians are testing a scale by using a weight known to weigh exactly 1000 grams. The standard deviation for scale reading is known to be σ = 2.3. They weigh this weight on the scale 49 times and read the result each time. The 49 scale readings have a sample mean of x = 999.0 grams.

The calibration point is set too low if the mean scale reading is less than 1000 grams.

Let [tex]\mu[/tex] = mean scale reading

So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \geq[/tex] 1000 grams     {means that the calibration point is not set too low}

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 1000 grams     {means that the calibration point is set too low}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

                         T.S. =  [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]  ~ N(0,1)

where, [tex]\bar X[/tex] = sample mean = 999 grams

            [tex]\sigma[/tex] = population standard deviation = 2.3 grams

            n = sample of scale readings = 49

So, test statistics  =  [tex]\frac{999-1000}{\frac{2.3}{\sqrt{49} } }[/tex]  

                              =  -3.04

The value of z test statistics is -3.04.

Now, P-value of the test statistics is given by the following formula;

         P-value = P(Z < -3.04) = 1 - P(Z [tex]\leq[/tex] 3.04)

                                              = 1 - 0.99882 = 0.00118

Since, the P-value is less than the level of significance as 0.01 > 0.00118, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the calibration point is set too low.

Step-by-step explanation:

A health inspector must visit 3 of 13 restaurants on Monday.

In how many way can she pick 3 restaurants = 13 C 4 !

nCr ! = [tex]\frac{n!}{r!(n-r)!}[/tex]

13 C 4! = [tex]\frac{13!}{3!(10)!}[/tex]

= [tex]\frac{11.12.13}{2.3}[/tex]

= 286

Health inspector can pick 3 restaurants of 13 restaurants on Monday in 286 ways

Final answer:

There are 286 different ways for a health inspector to pick 3 out of 13 restaurants to visit, calculated using the combination formula C(n, k) = n! / (k!(n - k)!).

Explanation:

The student asked how many different ways a health inspector could pick 3 out of 13 restaurants to visit on Monday. This is a problem of combinatorics, specifically concerning combinations without repetition since the order of choosing restaurants does not matter.

To calculate this, we use the combination formula, which is defined as:

C(n, k) = n! / (k!(n - k)!)

Where:

Applying the formula:

C(13, 3) = 13! / (3!(13 - 3)!) = 13! / (3! * 10!) = (13*12*11) / (3*2*1) = 286

So, the health inspector has 286 different ways to pick 3 restaurants from a list of 13.

Answer:

k = 37

Step-by-step explanation:

[tex]k = 13t - 2 \\ \therefore \: k = 13 \times 3 - 2 \\ \therefore \: k = 39 - 2 \\ \huge \red{ \boxed{ \therefore \: k = 37}}[/tex]

[tex]\text{Solve the equation to make it into slope intercept form}\\\\\text{Slope intercept form: y = mx + b}\\\\\text{Solve:}\\\\-2y=6(2x-2)\\\\\text{Distribute the 6 to the variables inside the parenthesis}\\\\-2y=12x-12\\\\\text{Divide both sides by -2}\\\\y=-6x+6\\\\\text{The slope would be -6 and the y intercept would be 6}\\\\\boxed{\text{Slope: -6 y-intercept: 6}}[/tex]

Answer:

Step-by-step explanation:

Hello!

The objective is to test if the population proportion of gamers that prefer consoles is less than 28% as the manufacturer claims.

Of 341 surveyed players, 89 said that they prefer using a console.

The sample resulting sample proportion is p'= 89/341= 0.26

If the company claims is true then p<0.28, this will be the alternative hypothesis of the test.

H₀: p ≥ 0.28

H₁: p < 0.28

α: 0.05

To study the population proportion you have to use the approximation of the standard normal [tex]Z= \frac{p'-p}{\sqrt{\frac{p(1-p)}{n} } }[/tex]≈N(0;1)

[tex]Z_{H_0}= \frac{0.26-0.28}{\sqrt{\frac{0.28*0.72}{341} } }= -0.82[/tex]

This test is one-tailed left, i.e. that you'll reject the null hypothesis to small values of Z, and so is the p-value, you can obtain it looking under the standard normal distribution for the probability of obtaining at most -0.82:

P(Z≤-0.82)= 0.206

Using the p-value approach:

If p-value ≤ α, reject the null hypothesis

If p-value > α, don't reject the null hypothesis

The decision is to not reject the null hypothesis.

Then at a level of 5%, you can conclude that the population proportion of gamers that prefer playing on consoles is at least 28%.

I hope this helps!

The correct hypotheses to be tested are:

H0: p = 0.5 (The proportion of songs loaded by Rina is 0.5, meaning Ed and Rina loaded an equal proportion of songs.)

Ha: p = 0.5 (The proportion of songs loaded by Rina is not equal to 0.5, meaning Ed and Rina loaded different proportions of songs.)

To determine how strong the evidence is that Ed and Rina have each loaded a different proportion of songs into the player, we can perform a hypothesis test. The test statistic for a proportion in a binomial setting is given by:

[tex]\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \][/tex]

where:

- [tex]\(\hat{p}\)[/tex] is the sample proportion of successes (in this case, the proportion of songs by Rina),

- [tex]\(p_0\)[/tex] is the hypothesized proportion under the null hypothesis (0.5),

- [tex]\(n\)[/tex] is the sample size (53 songs).

Given that 35 out of 53 songs were by Rina, we calculate [tex]\(\hat{p}\)[/tex] as:

[tex]\[ \hat{p} = \frac{35}{53} \approx 0.660 \][/tex]

Now we can calculate the test statistic [tex]\(z\)[/tex] :

[tex]\[ z = \frac{0.660 - 0.5}{\sqrt{\frac{0.5(1-0.5)}{53}}} = \frac{0.160}{\sqrt{\frac{0.25}{53}}} \approx \frac{0.160}{\sqrt{0.004717}} \approx \frac{0.160}{0.0686} \approx 2.33 \][/tex]

The P-value for a two-tailed test is the probability of observing a test statistic as extreme as or more extreme than the one observed, under the assumption that the null hypothesis is true. We can find the P-value by looking up the z-score in a standard normal distribution table or using a calculator:

[tex]\[ P\text{-value} = 2 \times P(Z > |2.33|) \approx 2 \times 0.0099 \approx 0.0198 \][/tex]

Since the P-value (0.0198) is less than the significance level of 0.05, we reject the null hypothesis in favour of the alternative hypothesis. This means there is strong evidence to suggest that Ed and Rina have each loaded a different proportion of songs into the player.

Before we conclude, we must check the conditions for the use of this test:

1. The samples are independent. Each song selection is made independently of the others.

2. The number of successes and failures are each at least 10. In this case, Rina's songs (35) and Ed's songs (18) are both greater than 10, so this condition is satisfied.

3. The sample size is less than 10% of the population size. The population size is 3476 songs, and the sample size is 53 songs, which is much less than 10% of the population, so this condition is also satisfied.

Since all conditions are met, we can confidently conclude that there is strong evidence that Ed and Rina have each loaded a different proportion of songs into the player.

Answer:

a) 0.3889

b) 0.5

c) 0.8333

d) The mean is 250 and the standard deviation is 51.96.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability of finding a value of X higher than x is:

[tex]P(X > x) = 1 - \frac{x - a}{b-a}[/tex]

The probability of finding a value of X between c and d is:

[tex]P(c \leq X \leq d) = \frac{d - c}{b - a}[/tex]

The mean and the standard deviation are, respectively:

[tex]M = \frac{a+b}{2}[/tex]

[tex]S = \sqrt{\frac{b-a}^{2}{12}}[/tex]

A random variable follows the continuous uniform distribution between 160 and 340.

This means that [tex]a = 160, b = 340[/tex]

a)

[tex]P(220 \leq X \leq 290) = \frac{290 - 220}{340 - 160} = 0.3889[/tex]

b)

[tex]P(160 \leq X \leq 250) = \frac{250 - 160}{340 - 160} = 0.5[/tex]

c)

[tex]P(X > 190) = 1 - \frac{190 - 160}{340 - 160} = 0.8333[/tex]

d)

[tex]M = \frac{160 + 340}{2} = 250[/tex]

[tex]S = \sqrt{\frac{340 - 160}^{2}{12}} = 51.96[/tex]

The mean is 250 and the standard deviation is 51.96.

Answer:

73/100

Step-by-step explanation:

.73 is 73 hundredth which is also 73/100.

Answer:

73/100

Step-by-step explanation:

.73

There are 2 numbers after the decimal, which means the denominator is 1 with 2 zeros after it

73/100

This is as simple as it get since 73 is a prime number

Answer:

a.

The random variable has binomial distribution with parameters p = 0.25, n=5.

b.

[tex]P(Y=2) = {5 \choose 2} (0.25)^2(0.75)^3 = 0.264\\P(Y<1 ) = P(Y = 0 ) = (0.75)^5 = 0.237[/tex]

Step-by-step explanation:

a.

Remember what a random variable with binomial distribution is. It is a random variable that counts the number of successes and in n trials of an experiment.

In this case, if you have  0,1,2,3,4,5  bottles of water, your success is that  the bottle has tap water, and the probability of that success is p = 0.25. and the number of trials is n = 5.

b.

Using the formula for the binomial distribution you get that

[tex]P(Y=2) = {5 \choose 2} (0.25)^2(0.75)^3 = 0.264\\P(Y<1 ) = P(Y = 0 ) = (0.75)^5 = 0.237[/tex]

Answer:the answer is (A)

Step-by-step explanation:

Because the survey face-to-face, one at a time, with an administrator of the school the students would likely reply that they do not text and drive even if they actually do. This response bias would cause the sample proportion to underestimate the value of the true proportion of all students at this school that text and drive.

The design of the survey contains response bias, which is likely to underestimate the true proportion of students who text while driving due to students' reluctance to admit to prohibited and risky behaviors to an authority figure.

The potential bias present in the design of this survey is response bias.

This type of bias occurs when the way a question is asked leads the respondents to give a certain answer

. In this case, by directly asking the students if they engage in the prohibited behavior of texting while driving, there is a likelihood that students may not answer truthfully due to fear of repercussions, the desire to give socially acceptable answers, or reluctance to admit to risky behaviors in a face-to-face setting with a school authority figure.

The likely direction of the bias is that it would underestimate the true proportion of students who text while driving since students might not want to admit to a behavior that is both dangerous and frowned upon while talking to a school administrator.

Answer:

  6

Step-by-step explanation:

Put the given point into the equation and solve for a.

  f(-2) = a(-2 +3)² -4 = a -4 = 2

  a = 6 . . . . . add 4

The a-value in the equation is 6.

Answer:

A.) 2 x minus 4 y = 8. x + y = 7.

Step-by-step explanation:

A.) 2 x minus 4 y = 8. x + y = 7.

2x-4y = 8

x+y=7

This has one solution

B.) 3 x + y = negative 1. 6 x + 2 y = negative 2.

3x+y = -1

Multiply by 2

6x +2y = -2

This is the second equation  so it has infinite solutions

C.) 3 x + 3 y = 3. Negative 6 x minus 6 y = 3.

3x+3y =3

Multiply by -2

-6x-6y = -6

The equation is the same but the constant, which means they are parallel lines, they have zero solutions

D.) 3 x minus 2 y = 4. Negative 3 x + 2 y = 4.

3x -2y = 4

Multiply by -1

-3x +2y = -4

The equation is the same but the constant, which means they are parallel lines, they have zero solutions

Answer:

A.) 2 x minus 4 y = 8. x + y = 7.

Step-by-step explanation:

A.) 2 x minus 4 y = 8. x + y = 7.

2x - 4y = 8

x - 2y = 4

x = 4 + 2y

x + y = 7

4 + 2y + y = 7

3y = 3

y = 1

x = 4 + 2

x = 6

Only solution: (6,1)

Answer:

Contribution

Step-by-step explanation:

The verb would be used as:

I like to contribute to discussions.

The noun would be used as:

My contribution to the discussion was alright.

I hope this helped! (Sorry for the dry examples, I couldn't think of anything else)

Answer:

The population consists of the 201 people who bought tires last month.

The sample consists of the 51 people who responded.

Step-by-step explanation:

In statistical analysis, the population is a set of all values that can be considered for an experiment. It is a set of observations from which a sample is selected. The sample is a subset of the population.

In this case, it is provided that a tire company sends an email survey to all customers who purchase new tires each month.

The number of tires sold last month was N = 201.

The email survey was sent to 100 of these 201 people and 51 responded.

In this case, since the survey is done every month, the population consists of all the people who bought tires the previous month.

So, the population consists of the 201 people who bought tires last month.

The number of people who responded was, n = 51.

The sample consists of these 51 people because the data from these 51 people would be used to conduct and draw conclusion about the survey.

So, the sample consists of the 51 people who responded.

Answer:

By the Central Limit Theorem, the mean of the average scores you get will be close to 1012.5.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean score of the population:

1012.5

If you do this many times, the mean of the average scores you get will be close to

By the Central Limit Theorem, the mean of the average scores you get will be close to 1012.5.

By definition, if [tex]\vec v[/tex] is an eigenvector of [tex]\mathbf A[/tex] (with associated eigenvalue 8), then

[tex]\mathbf A\vec v=8\vec v[/tex]

[tex]\mathbf A^4\vec v=\mathbf A^3\cdot\mathbf A\vec v=8\mathbf A^3\vec v[/tex]

Rinse and repeat to find

[tex]\mathbf A^4\vec v=8^4\vec v[/tex]

so that [tex]\mathbf A^4[/tex] has a corresponding eigenvalue of 4096.

[tex]\mathbf A^{-1}\vec v=\lambda\vec v\implies\mathbf A\cdot\mathbf A^{-1}\vec v=\mathbf A\cdot\lambda \vec v\implies\vec v=8\lambda\vec v[/tex]

For this to hold, we require [tex]8\lambda=1[/tex], or [tex]\lambda=\frac18[/tex]. So [tex]\mathbf A^{-1}[/tex] has a corresonding eigenvalue of 1/8.

[tex](\mathbf A+4\mathbf I_n)\vec v=\mathbf A\vec v+4\mathbf I_n\vec v=8\vec v+4\vec v=12\vec v[/tex]

so that [tex]\mathbf A+4\mathbf I_n[/tex] has an associated eigenvalue of 12.

[tex]5\mathbf A\vec v=5\cdot8\vec v=40\vec v[/tex]

so the eigenvalue of [tex]5\mathbf A[/tex] is 40.

The matrix [tex]A^{4}[/tex]   has eigen value of 4096.

The matrix [tex]A^{-1}[/tex] has eigen value of [tex]\frac{1}{8}[/tex]

The eigen value of (A+4I) is 12.

The eigen value of [tex]5Av[/tex] is 40.

Eigenvalues are the set of scalar values that is associated with the set of linear equations most probably in the matrix equations.

The eigenvectors are also termed as characteristic roots.

If v is an eigen vector of matrix A associated with eigen value [tex]\lambda[/tex].

   Then,         [tex]Av=\lambda v[/tex]

it is given that, v is an eigen vector of A with eign value 8.

so taht,  [tex]Av=8v[/tex]

We have to find eigen value for [tex]A^{4}[/tex].

 [tex]A^{4}v=A^{3}(8v)=8^{4}v=4096v[/tex]

Thus, [tex]A^{4}[/tex] has eigen value of 4096.

If A has eigen value of 8 . Then [tex]A^{-1}[/tex] has eigen value of [tex]\frac{1}{8}[/tex]

The eigen value of (A+4I) is computed as,

        [tex](A+4I_{n})v=Av+4v=8v+4v=12v[/tex]

The eigen value of (A+4I) is 12.

The eigen value of [tex]5Av[/tex] is,

       [tex]5Av=5*8v=40v[/tex]

The eigen value of [tex]5Av[/tex] is 40.

Learn more about the eigen vectors here:

https://brainly.com/question/15423383

Given:

The height of the the tree from where the tree was broken = 13 ft

The distance from the foot of the tree to the broken top = 36 ft

To find the height of the tree.

Formula

By Pythagoras theorem we get,

h² = l²+b²,

where, h be the hypotenuse

b be the base and

l be the other side of the triangle along the right angle.

Now,

Putting, l =13 and b = 36 we get,

[tex]h^{2} =36^{2} +13^{2}[/tex]

or, [tex]h^{2} = 1296+169[/tex]

or, [tex]h^{2} =1465[/tex]

or, [tex]h=\sqrt{1465}[/tex]

or, [tex]h = 38.3[/tex]

Therefore,

The height of the tree is about = 13+38.3 ft = 51.3 ft

Hence,

The height of the tree is about 51.3 ft.