The regression equation Credits = 17.9 − 0.09 Work was estimated from a sample of 21 statistics students. Credits is the number of college credits taken and Work is the number of hours worked per week at an outside job. (Round your answers to 1 decimal place.) (a) Choose the correct statement. An increase in the number of hours worked per week increases the expected number of credits. A decrease in the number of hours worked per week decreases the expected number of credits. An increase in the number of hours worked per week decreases the expected number of credits. (b) Choose the right option. The intercept is not meaningful as a student who works outside may not earn any credits. The intercept is meaningful as a student may not have a job outside of school. (c) If Work = 0, then Credits = . If Work = 60, then Credits = .

The equation in slope-intercept from is [tex]y = \frac{9}{8}x - \frac{3}{8}[/tex].

Step-by-step explanation:

Given that,

x = 8t+3

y = 9t+ 3

Now, getting the value of t on both side to compare the equations.

[tex]x = 8t+3[/tex]

[tex]x -3 = 8t[/tex]

[tex]t = \frac{x-3}{8}[/tex]

also

[tex]y = 9t+ 3[/tex]

[tex]t= \frac{y-3}{9}[/tex]

Now, by comparing the values of t, we get

[tex]\frac{x-3}{8} = \frac{y-3}{9}[/tex]

[tex]9x - 27 = 8y -24[/tex]

[tex]9x - 3 = 8y[/tex]

[tex]y = \frac{9}{8}x - \frac{3}{8}[/tex]

Hence, the equation in slope-intercept from is [tex]y = \frac{9}{8}x - \frac{3}{8}[/tex].

Answer:

86.96

Step-by-step explanation:

not sure what is height or base.

Answer:

86.96

Step-by-step explanation:

Answer:

x = -7

Step-by-step explanation:

2x + 3 = x - 4

→ Minus x from both sides to isolate -4

x + 3 = -4

→ Minus 3 from both sides to isolate x and henceforth find the value of x

x = -7

Answer:

It equals 2/5.

Step-by-step explanation:

Basically you ignore the 5 because it stays the same. All you have to add are the numerators. 1+1=2. So it would be 2/5.

A scenario:

2 pizzas cut into 5th's. Everyone ate 4 in each and kept one slice. When you add those 2 together you would be 2 slices.

Answer:

Step-by-step explanation:

1/8

1/2 multiply 1/4

Answer:

243

Step-by-step explanation:

3x3=9

9x3=27

27x3=81

81x3=243

Answer:

1. Sample Mean = 2.50

2. Sample Variance = 1.35

3. Standard Deviation = 1.16

Step-by-step explanation:

Note: As this question is not complete, similar can be found on internet where following are asked to calculate and here I will calculate these as well.

1. Sample Mean.

2. Sample Variance.

3. Sample Standard Deviation.

So, for sample mean to calculate from the given data points. We need to apply to following formula to calculate sample mean.

Sample Mean = (X1 + X2 + .... + XN) divided by Total number of data points.

Sample Mean = (0.17 + 1.94 + 2.62 + 2.35 + 3.05 + 3.15 + 2.53 + 4.81 + 1.92) divided by (9).

Sample Mean = 2.50

Likewise, from the answer of sample mean we can calculate sample variance by following steps.

Solution:

1.      Subtract the obtained mean from each of the data point given.

       (0.17 - 2.50) = -2.33

       (1.94 - 2.50) = -0.56

       (2.62 - 2.50) = 0.12

       (2.35 - 2.50) = -0.15

       (3.05 - 2.50) = 0.55

       (3.15 - 2.50) = 0.65

       (2.53 - 2.50) = 0.03

       (4.81 - 2.50) =  2.31

       (1.92 - 2.50) = -0.58

2.      Square each of the differences obtained in step 1.

            [tex]-2.33^{2}[/tex] = 5.43

            [tex]-0.56^{2}[/tex] = 0.31

               [tex]0.12^{2}[/tex] = 0.01

            [tex]-0.15^{2}[/tex] = 0.02

              [tex]0.55^{2}[/tex]  = 0.30

              [tex]0.65^{2}[/tex]  = 0.42

              [tex]0.03^{2}[/tex] = 0.01

              [tex]2.31^{2}[/tex] = 5.34    

           [tex]-0.58^{2}[/tex] = 0.34    

3.  Sum all of these squares.

(5.43 + 0.31 + 0.01 + 0.02 + 0.30 + 0.42 + 0.01 + 5.34 + 0.34) = 12.18

4. Divide 12.18 by (n-1), where n = 9.

Sample Variance =  12.18/9 = 1.35

Now, by using sample variance, we can calculate standard deviation with easy simple steps.

So, in order to find standard deviation, we just need to take the square root of sample variance that we have already calculated above.

Standard Deviation = [tex]\sqrt{1.35}[/tex] = 1.16

Good Luck!

Answer:

  1

Step-by-step explanation:

You "complete the square" by adding the square of half the x-term coefficient. Here, that is ...

 ((-2)/2)² = 1 . . . . value added to complete the square

If you want to keep 0 on the right, you must also subtract this value:

  x² -2x -36 = 0

  x² -2x +1 -36 -1 = 0 . . . . . . add and subtract 1 on the left

  (x -1)² -37 = 0 . . . . . . . . . . . written as a square

Given Information:

Population = n = 800

Probability = p = 68% = 0.68

Answer:

We can say with 68% confidence that 568 lies in the range of (518, 570) therefore, it would not be unusual to get 568 consumers who recognize the Dull Computer Company name.

Step-by-step explanation:

Let us first find out the mean and standard deviation.

mean = μ = np

μ = 800*0.68

μ = 544

standard deviation = σ = √np(1-p)

σ = √800*0.68(1-0.68)

σ = 13.2

we know that 68% of data fall within 2 standard deviations from the mean

μ ± 2σ = 544-2*13.2, 544+2*13.2

μ ± 2σ = 544-26.4 , 544+26.4

μ ± 2σ =  517.6, 570.4

μ ± 2σ = 518, 570

We can say with 68% confidence that 568 lies in the range of (518, 570) therefore, it would not be unusual to get 568 consumers who recognize the Dull Computer Company name.

Answer:

y = x+1

Step-by-step explanation:

You should use the point-slope form of y-y1 = m(x-x1) to solve this.

First you use the slope formula to find your slope. (y2-y1)/(x2-x1)

(-1 - -3)/(-2 - -4) = 2/2 = 1

With the slope of 1 found, just plug a point into the point-slope form.

y-(-1) = (1)(x-(-2)) ->

y+1 = x+2 ->

y = x+1

y=x+1 is the equation of a line that passes through (-2, -1) and (-4, -3)

The slope of the line is the ratio of the rise to the run, or rise divided by the run. It describes the steepness of line in the coordinate plane.

The slope intercept form of a line is y=mx+b, where m is slope and b is the y intercept.

The slope of line passing through two points (x₁, y₁) and (x₂, y₂) is

m=y₂-y₁/x₂-x₁

The slope of line passing through  (-2, -1) and (-4, -3)

m = -3+1/-4+2

=-2/-2= 1

Now let us find y intercept

-1=1(-2)+b

-1=-2+b

b=1

Hence, y=x+1 is the equation of a line that passes through (-2, -1) and (-4, -3)

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Answer:

We can select a lower confidence level and increase the sample size.

Step-by-step explanation:

The precision of the confidence interval depends on the margin of error ME = Zcritical * Sqrt[(p(1-p)/n]

In this Zcritical value is in the numerator. Z critical decreases as Confidence level decreases. (Zc for 99% = 2.576, Zc for 95% is 1.96, Zc for 90% = 1.645). Therefore decreasing the Confidence level decreases ME.

Also we see that sample size n is in the denominator. So the ME decreases as sample size increases.

Therefore, We can select a lower confidence level and increase the sample size.

The best option that could be used to produce higher precision in our estimates of the population proportion is;

Option C; We can select a lower confidence level and increase the sample size.

Formula for confidence interval is given as;

CI = p^ ± z√(p^(1 - p^)/n)

Where;

p^ is the sample proportion

z is the critical value at given confidence level

n is the sample size

Now, the margin of error from the CI formula is:

MOE = z√(p^(1 - p^)/n)

Now, the lesser the margin of error, the narrower the confidence interval and thus the more precise is the estimate of the population proportion.

Now, looking at the formula for MOE, two things that could change aside the proportion is;

z and n.

Now, the possible values of z are;

At CL of 99%; z = 2.576

At CL of 95%; z = 1.96

At CL of 90%; z = 1.645

We can see that the higher the confidence level, the higher the critical value and Invariably the higher the MOE.

Thus, to have a narrow CI, we need to use a lower value of CL and increase the sample size.

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Answer:

Length of the chord: 24 units

Angles are equal

Step-by-step explanation:

Drop a Perpendicular from the centre onto the chord. It will divide the chord into two equal parts, d units each

d² + 16² = r²

d² = 20² - 16²

d² = 144

d = 12

Chord = 2d = 24 units

Measure of central angle and the corresponding chord related to the measure of the arc intercepted by the chord are the same, they're equal

Answer:

The entire chord length is 12*2 = 24

The degree measure of a minor arc is equal to the measure of the central angle that intercepts it.

Step-by-step explanation:

We can make a right triangle to solve for 1/2 of the chord length.  The hypotenuse is 20 and one of the legs is 16

a^2+b^2 = c^2

16^2 + b^2 = 20^2

256 +b^2 = 400

Subtract 256 from each side

b^2 = 400-256

b^2 =144

Take the square root of each side

b = 12

That means 1/2 of the chord length is 12

The entire chord length is 12*2 = 24

A sample size of 188 would be needed to achieve a 6% margin of error at a 90% confidence level.

Used the formula for sample size calculation:

[tex]n = \dfrac{z^2 \times p (1 - p)}{E^2}[/tex]

Where: n = sample size.

Z = Z-score corresponding to the desired confidence level.

p = estimated proportion of the population that supports the candidate.

E = desired margin of error.

Here we have to give that,

90% confidence level corresponds to a Z-score of approximately 1.645.

That is, z = 1.645

p = 0.5

E = 0.06

Substituting the given values into the formula, we get:

[tex]n = \dfrac{z^2 \times p (1 - p)}{E^2}[/tex]

[tex]n = \dfrac{(1.645)^2 \times 0.5(1 - 0.5)}{0.06^2}[/tex]

[tex]n = \dfrac{0.6765}{0.0036}[/tex]

[tex]n =187.9[/tex]

Rounded to whole numbers,

[tex]n = 188[/tex]

Therefore, the size of the sample would be 188.

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In order to achieve a 6% margin of error and a 90% confidence level in a poll, the correct sample size must be determined using the error bound formula. A larger and representative sample will minimize the margin of error and improve the poll's precision and accuracy. A number of key factors, like identification of the target population and select a random sample, should be considered while conducting a poll.

The subject matter dealt with here involves conducting a poll to gauge public opinion, with a desired accuracy defined by a 6% margin of error and a 90% confidence level. This is primarily related to Statistics, a branch of Mathematics.

To determine an appropriate sample size for a given margin of error and confidence level, the error bound formula is utilized. Although the actual formula could appear complex for some, many online calculators can complete the task. The 90% confidence level means that if we were to take multiple sets of samples, roughly 90% of the calculated confidence intervals would contain the true population mean or proportion.

Furthermore, sample size influences the margin of error; a larger sample reduces the margin of error and makes the poll result more precise and accurate. As a rule of thumb, the sample should be random and representative of the population in question to ensure the accuracy and reliability of poll results.

Take note that many criteria must be met to conduct a valid and scientifically accurate poll, some of which include: identifying the desired population of respondents, ensuring the sample is both random and representative of the wider population, and interviewing a sufficient number of citizens to achieve a reasonable sample size.

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Answer:

Yes, it is convincing evidence to conclude that the proportion of red cars that drive too fast on this

highway is greater than the proportion of non-red cars that drive too fast.

Step-by-step explanation:

From the, we wish to first test;

H0; P_r - P_o = 0

And; H0; P_r - P_o > 0

Where; P_r and P_o are the proportion of red cars and other cars, respectively, who are driving too fast.

We will use a significance level of a = 0.05.

Thus;

The procedure is a two-sample z-test for the difference of proportions.

For, Random Conditions: The policemen chose cars randomly by rolling a die.

10%: We can safely assume that the number of cars driving past the rest area is essentially infinite, so the 10% restriction does not apply.

Large counts: The number of successes and failures in the two groups are 18, 10, 75, and 130—all of which are at least 10.

So, P_r = 18/28 = 0.64

P_o = 75/205 = 0.37

P_c = (18 + 75)/(28 + 205) = 0.4

Thus:

z = [(0.64 - 0.37) - 0]/√[[(0.4 x 0.6)/28] + [(0.4 x 0.6)/205]]

z = 2.73

From the one tailed z-score calculator online, I got P value = 0.003167

Thus, the P-value of 0.0032 is less than a = 0.05, so we reject H0. We

have sufficient evidence to conclude that the proportion of red cars that drive too fast on this

highway is greater than the proportion of non-red cars that drive too fast.

QUESTION BEGINNING

Given a snail is traveling along a straight path. The snail’s velocity can be modeled by [tex]v(t)=1.4ln(1+t^2)[/tex] inches per minute for 0 ≤ t ≤ 15 minutes.

Answer:

B=22.35 Inches per minutes

Step-by-step explanation:

If the snail's velocity is [tex]v(t)=1.4ln(1+t^2)[/tex] per minute, its displacement for             0 ≤ t ≤ 15 minutes is given by the integral:

[tex]\int v(t) dt=\int (1.4ln(1+t^2))dt=76.04307[/tex]

The constant acceleration of the ant is 2 Inches per minute.

The velocity of the ant therefore, twill be:

[tex]\int 2 dt=2t+K, $where K is a constant of integration$[/tex]

For the interval, 12≤t≤15, the displacement of the ant is:

[tex]\int_{12}^{15}(2t+K) dt=81+3K[/tex]

Since the snails displacement and that of the ant are equal in 12≤t≤15.

81+3K=76.04307

3K=76.04307-81

3K=-4.95693

K=-1.65231

At t=12, the velocity of the ant  is therefore:

2t+K=2(12)-1.65231=22.348 Inches per minutes

B=22.348 Inches per minutes

Final answer:

The question examines a scenario involving an ant accelerating to catch up with a snail, focusing on determining the ant's velocity at the catch-up point. Utilizing the principle of kinematics and acceleration, the answer lies in relating the acceleration to the final velocity, dependent on the time variable, which underscores the direct relationship between acceleration and velocity increase.

Explanation:

The question essentially deals with kinematics, specifically with the scenario of an ant accelerating from rest to catch up with a snail. Given the ant accelerates at 2 inches per minute per minute, and eventually matches velocities with the snail, the goal is to find the ant's velocity (B inches per minute) at the moment it catches up.

Firstly, we know the acceleration (a) is 2 inches/minute2. From kinematic equations, specifically v = u + at, where 'u' is the initial velocity (0 in this case since the ant starts from rest), 'a' is acceleration, and 't' is time, we can plug in our values. The time variable 't' here represents the duration from when the ant starts until it catches up with the snail, which is not explicitly given but is critical in understanding the rate of acceleration's contribution to velocity.

For the purpose of solving this problem, without explicit time or distances, we focus on the principle that the ant's final velocity (B) is a product of its acceleration over a given period: B = 0 + (2)t. Thus, the value of 'B' entirely depends on the time 't', demonstrating a direct correlation between acceleration duration and velocity increase.

Answer:

a) 2.81 b)-2.33 c) -2.88 d)3.09

Step-by-step explanation:

The complete question is:

Determine the value of z so that the area under the standard normal curve

a. in the right tail is 0.025 Round your answer to two decimal places.

b. in the left tail is 0.01 Round your answer to two decimal places.

c. in the left tail is 0.002 Round your answer to two decimal places.

d. in the right tail is 0.01 Round your answer to two decimal places.

a)P( Z> ???)=  0.0025

P(Z> ???)= 1-P(Z<???)

P(Z<???)= 1-0.0025

P(Z<??)= 0.9975

From Z distribution table,

Z = 2.81

b) P(Z<???)= 0.01

From Z distribution table

Z= -2.33

c) P(Z< ??? ) = 0.002

From Z distribution table

Z= -2.88

d) P( Z> ???)=  0.001

P(Z> ???)= 1-P(Z<???)

P(Z<???)= 1-0.001

P(Z<??)= 0.999

From Z distribution table,

Z=3.09

Answer:

2,4,5

Step-by-step explanation:

Answer:2,4,and5

Step-by-step explanation:Common sense

Answer:

=2p(-2p-5)

Step-by-step explanation:

2p(–3p2 + 4p – 5)

=2p(4p-6p-5)

=2p(-2p-5)

Answer:

Step-by-step explanation:

REcall the following definition of induced operation.

Let * be a binary operation over a set S and H a subset of S. If for every a,b elements in H it happens that a*b is also in H, then the binary operation that is obtained by restricting * to H is called the induced operation.

So, according to this definition, we must show that given two matrices of the specific subset, the product is also in the subset.

For this problem, recall this property of the determinant. Given A,B matrices in Mn(R) then det(AB) = det(A)*det(B).

Case SL2(R):

Let A,B matrices in SL2(R). Then, det(A) and det(B) is different from zero. So

[tex]\text{det(AB)} = \text{det}(A)\text{det}(B)\neq 0[/tex].

So AB is also in SL2(R).

Case GL2(R):

Let A,B matrices in GL2(R). Then, det(A)= det(B)=1 is different from zero. So

[tex]\text{det(AB)} = \text{det}(A)\text{det}(B)=1\cdot 1 = 1[/tex].

So AB is also in GL2(R).

With these, we have proved that the matrix multiplication over SL2(R) and GL2(R) is an induced operation from the matrix multiplication over M2(R).

Answer:

Step-by-step explanation:

Zip codes with 5 digits (all five digits can be repeated): 10^5=100,000 if first digit is 0, its no longer 5-digit right? Zip codes with no digit being repeated: 10*9*8*7*6=30,240 if first digit is 0, its no longer 5-digit right? Zip codes with at least one digit being repeated: 90000-27216 = 62784 ?N

Answer:

It is A,B,and E.

I just did the question and it was right

The true statements about the dot plots are (A) Both have the same number of data points, (B) Both means are between 3 and 4. and (C) Westwood has less variability than Middleton.

Mean of a set of data is defined as the average of all the values. It gives the exact middle point of the data set.

Given are two plots, Middleton and Westwood.

Data points in plot 1 are 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 9

Total number of data points in plot 1 or Middleton = 20

Data points in plot 2 are 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6

Total number of data points in plot 2 or Westwood = 20

So, both have the same number of data points.

Mean of Middleton = Sum of the data points / Total number of data points

                                = [1 + (4×2) + (5×3) + (4×4) + (4×5) + 6 + 9] / 20

                                = 75/20

                                = 3.75

Similarly,

Mean of Westwood = 77/20 = 3.85

So both means are between 3 and 4.

Median is the exact middle element when arranged in increasing or decreasing order.

Exact middle point is the average of 10th and 11th element.

Median of Middleton = (3 + 4)/2 = 3.5

Median of Westwood = (4 + 4)/2 = 4

Both have different medians.

Range of the data set is the difference of maximum point and minimum point.

Range of Middleton = 9 - 1 = 8

Range of Westwood = 6 - 2 = 4

Both have different ranges.

Range is a measure of variation.

Range of Westwood is less than range of Middleton.

So, Westwood has less variability than Middleton.

Hence the true statements are A, B and C.

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Answer:

The critical value of t at 0.01 level of significance is 2.66.

Step-by-step explanation:

The hypothesis for the two-tailed population mean can be defined as:

H₀: μ = μ₀ vs. H₀: μ ≠ μ₀

It is provided that the population standard deviation is not known.

Since there is no information about the population standard deviation, we will use a t-test for single mean.

The test statistic is defined as follows:

[tex]t_{cal.}=\frac{\bar x-\mu}{s/\sqrt{n}}\sim t_{\alpha/2, (n-1)}[/tex]

The information given is:

n = 55

α = 0.01

Compute the critical value of t as follows:

[tex]t_{\alpha/2, (n-1)}= t_{0.01/2, (55-1)}=t_{0.005, 54}=2.66[/tex]

*Use a t-table for the value.

If the desired degrees of freedom are not provided consider he next highest degree of freedom.

Thus, the critical value of t at 0.01 level of significance is 2.66.

Answer:

2nd, 4th and last one are correct answers

Step-by-step explanation:

[tex] {16}^{ \frac{5}{2} } \\ = ( {4}^{2} )^{ \frac{5}{2} } \\ = {4}^{{2} \times \frac{5}{2} } \\ = {4}^{5} \\ [/tex]

Options: 2nd, 4th and last one are correct answers

Answer:

200 pencils and 120 pens

Step-by-step explanation:

Let the number of pencils be x while pens be y.

Considering the cost of whole purchase then, we get the equation

0.5x+y=220

Considering the number of items bought, then the equation is

x-y=80

Adding the two equations then we have

1.5x=300

x=300/1.5=200

Consideeing that x-y=80 and x is 200 then

200-y=80

y=200-80=120

Therefore, the pencils were 200 and pens 120 pieces

Answer:

The proportion of chips that do not fail in the first 1000 hours of their use is 52%.

Step-by-step explanation:

The claim made by the company is that 52% of the chips do not fail in the first 1000 hours of their use.

A quality control manager wants to test the claim.

A one-proportion z-test can be used to determine whether the proportion of chips do not fail in the first 1000 hours of their use is 52% or not.

The hypothesis can be defined as:

H₀: The proportion of chips do not fail in the first 1000 hours of their use is 52%, i.e. p = 0.52.

Hₐ: The proportion of chips do not fail in the first 1000 hours of their use is different from 52%, i.e. p ≠ 0.52.

The information provided is:

[tex]\hat p=0.49\\n=900\\\alpha =0.02[/tex]

The test statistic value is:

[tex]z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.49-0.52}{\sqrt{\frac{0.52(1-0.52)}{900}}}=-1.80[/tex]

The test statistic value is -1.80.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.

Compute the p-value as follows:

[tex]p-value=2\times P(Z<-1.80)\\=2\times 0.03593\\=0.07186[/tex]

*Use a z-table.

The p-value = 0.07186 > α = 0.02.

The null hypothesis was failed to be rejected at 2% level of significance.

Conclusion:

The proportion of chips that do not fail in the first 1000 hours of their use is 52%.

Answer:

the answer is 1,3,4,5

Step-by-step explanation:

Answer:

We need to conduct a hypothesis in order to check if the mean the actual mean length of movies released during 2010-2015 is more than the data base value (a;ternative hypothesis) ,so then the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq 117[/tex]  

Alternative hypothesis:[tex]\mu > 117[/tex]  

Step-by-step explanation:

Data given and notation  

[tex]\bar X=118.44[/tex] represent the sample mean

[tex]s=8.82[/tex] represent the sample standard deviation

[tex]n=160[/tex] sample size  

[tex]\mu_o =117[/tex] represent the value that we want to test

[tex]\alpha[/tex] represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean the actual mean length of movies released during 2010-2015 is more than the data base value (a;ternative hypothesis) ,so then the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq 117[/tex]  

Alternative hypothesis:[tex]\mu > 117[/tex]  

Answer:

it is the  midpoint  formula because that is the only equation of  that is being divided by 2. The formula is X1 +X2 divied by 2 then it is Y1 + Y2 dived by two which gives you half of something like a line

Step-by-step explanation:

Answer:

mid point

Step-by-step explanation:

Step-by-step explanation:

[tex]1)\:v = 2 {e}^{3t} + 5 {e}^{ - 3t} \\ differentiating \: w.r.t.t \: on \: both \: sides \\ acceleration =\\ \frac{dv}{dt} = \frac{1}{dt} (2 {e}^{3t} + 5 {e}^{ - 3t} ) \\ = 2 \times \frac{1}{dt} {e}^{3t} + 5 \times \frac{1}{dt} {e}^{ - 3t} \\ \\ = 2 \times {e}^{3t} \times 3 + 5 \times {e}^{ - 3t} \times ( - 3) \\ = 6{e}^{3t} - 15{e}^{ - 3t} \\ \therefore \frac{dv}{dt} = 6{e}^{3t} - 15{e}^{ - 3t} \\ \therefore \bigg(\frac{dv}{dt} \bigg) _{t=1} = 6 {e}^{3 \times 1} - 15 {e}^ { - 3 \times 1} \\ \bigg(\frac{dv}{dt} \bigg) _{t=1} = 6 {e}^{3} - 15 {e}^ { - 3 } \\ acceleration = \\ \purple{ \boxed{ \bold{\bigg(\frac{dv}{dt} \bigg) _{t=1} = \bigg(\frac{6 {e}^{6} - 15}{ {e}^{3} } \bigg) \: m {s}^{ - 2} }}} \\ \\ 2) \: let \:s \: be \: the \: total \: distance \: travelled \\ \therefore \: s = v \times t \\ \therefore \: s= (2 {e}^{3t} + 5 {e}^{ - 3t}) \times t \\ \therefore \: (s)_{t=2} = (2 {e}^{3 \times 2} + 5 {e}^{ - 3 \times 2}) \times 2 \\ \therefore \: (s)_{t=2} = (2 {e}^{6} + 5 {e}^{ - 6}) \times 2 \\ \therefore \: (s)_{t=2} = 4 {e}^{6} + 10{e}^{ - 6} \\ \red{ \boxed{ \bold{\therefore \: (s)_{t=2} = \bigg(\frac{4 {e}^{12} + 10}{{e}^{ 6}} \: \bigg)m}}}\\ [/tex]

Answer:

The sampling distribution of the sample mean is:

[tex]\bar X\sim N(\mu_{\bar x}=2.4,\ \sigma_{\bar x}=0.40)[/tex]

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Then, the mean of the distribution of sample means is given by,

[tex]\mu_{\bar x}=\mu[/tex]

And the standard deviation of the distribution of sample means is given by,

[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}[/tex]

Let X = number of cars running a red light in a day, at a given intersection.

The information provided is:

[tex]E(X)=\mu=2.4\\SD(X)=\sigma=4\\n=100[/tex]

The sample selected is quite large, i.e. n = 100 > 30.

The Central limit theorem can be used to approximate the sampling distribution of the sample mean number of cars running a red light in a day, by the Normal distribution.

The mean of the sampling distribution of the sample mean is:

[tex]\mu_{\bar x}=\mu=2.4[/tex]

The standard deviation of the sampling distribution of the sample mean is:

[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{4}{\sqrt{100}}=0.40[/tex]

The sampling distribution of the sample mean is:

[tex]\bar X\sim N(\mu_{\bar x}=2.4,\ \sigma_{\bar x}=0.40)[/tex]

Final answer:

The sampling distribution of the sample mean is approximately normally distributed, with a mean of 2.4 cars and a standard deviation of 0.4 cars.

Explanation:

The sampling distribution of the sample mean can be described as follows: