A study is run to estimate the mean total cholesterol level in children 2 to 6 years of age. A sample of 9 participants is selected and their total cholesterol levels are measured as follows. 180 220 240 196 175 183 195 140 200 What is the sample mean?A)196B)200C)192.11D)180

Answer:100 people

Step-by-step explanation:

140 people x 5 ladle/1 person = 700 ladles full

700 ladles x 1 person/7 ladles = 100 people fed

Answer: Top Right, A rectangle has all the properties of a square.

Step-by-step explanation: Nobody how hard you try to make a rectangle a square, it won't work. A rectangle cannot have 4 equal slides.

Answer:

The mean is zero and the standard deviation is one.

Correct, since the sample mean is not always defined by a distribution with mean 0 and deviation 1. Since the normal standard distribution is not always representative of the sample mean dsitribution

Step-by-step explanation:

For this case we want to analyze the distribution of the sample mean given by:

[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

Let's analyze one by one the possible options:

The distribution of the values is obtained by repeated sampling.

False, for this case we assume that the data of each observation [tex] X_i[/tex] , [tex] i =1,2,...,n[/tex] is obtained from a repated sampling method

The samples are all of size n.

False, for this case we assume that the sample mean is obtained from n data values

The samples are all drawn from the same population.

False, for this case we assume that the data comes from the same distribution

The mean is zero and the standard deviation is one.

Correct, since the sample mean is not always defined by a distribution with mean 0 and deviation 1. Since the normal standard distribution is not always representative of the sample mean dsitribution

The mean is zero and the standard deviation is one.

It is not characteristic of the sampling distribution of a sample mean.

In the Sampling Distribution of the Sample Mean , If repeated random samples of a given size n are taken from a population .where the population mean is μ  and the population standard deviation is σ .then the mean of all sample means  is population mean μ .

Characteristic of Sampling Distribution ,

So, The mean is zero and the standard deviation is one in sampling distribution is not any mandatory condition.

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Answer:

5

Step-by-step explanation:

It be 5 because its in the middle

3,4,7,8

Step-by-step explanation:

1 2 3 4 5 6 7 8 9

3 4 7 8

Answer:

  (1000 m)/(1 km) and (60 min)/(1 h)

Step-by-step explanation:

Put the unit you don't want in a position to cancel the unit in the given number. Write the fraction so that the equivalent amount of the unit you do want is on the other side of the fraction bar.

Here we have km/min with km in the numerator. To cancel that, we need a fraction with km in the denominator. We want meters (m) in the numerator, so we need a fraction that has a number of meters equivalent to 1 km. That will be ...

  (1000 m)/(1 km)

This is one of the conversion factors we will need to multiply by.

__

We also have "min" in the denominator. To cancel that, we need a conversion factor with min in the numerator. The unit we want in the denominator is h (hours), so we need an equivalent for hours and minutes. That would be 60 min = 1 h, so we write the conversion factor as ...

  (60 min)/(1 h)

So, our conversion factors are ...

  (1000 m)/(1 km) and (60 min)/(1 h)

_____

The converted number is ...

  (4 km/min)(1000 m/km)(60 min/h) = 240,000 m/h

__

Comment on conversion factors

As you can see, we write the fraction so equal amounts are in numerator and denominator. Since the amounts are equal, the value of the fraction is 1. Multiplying by 1 in this form doesn't change the original value, it only changes the units.

Answer:

x = 8, x = -8

Step-by-step explanation:

Hello!

We can use the Difference of Squares Formula: [tex](a + b)(a - b) = a^2 - b^2[/tex]

We need the two terms to be perfect squares for this formula to work, and subtraction has to be the operation. As you can see x² is a square of x, and 64 is the square of 8.

Solve:

Set each factor to zero. (Zero Product Property)

And

The solutions are x = 8, and x = -8.

c^2 = a^2 + b^2 - 2(ab)(cos C)

c^2 + 2(ab)(cos C) = a^2 + b^2

2(ab)(cos C) = a^2 + b^2 - c^2

cos C = (a^2 + b^2 - c^2) / 2ab - Answer choice E

Hope this helps! :)

To make COS the subject of a formula, the equation is rearranged such that COS is isolated. An example can be in the equation a = b cos(x) which can be rearranged as cos(x) = a/b. A complete formula is necessary for an accurate step-by-step guide.

To make COS the subject of a formula, it typically involves other known quantities represented by variables and constants. For example, in the equation a = b cos(x), we can make cos(x) the subject of the equation by rearranging it to: cos(x) = a/b. However, to provide a more accurate step-by-step guide, the complete formula is necessary. This principle can be applied to various trigonometric formulas so that COS becomes the main focus.

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Answer:

(a) P (Z < 2.36) = 0.9909                    (b) P (Z > 2.36) = 0.0091

(c) P (Z < -1.22) = 0.1112                      (d) P (1.13 < Z > 3.35)  = 0.1288

(e) P (-0.77< Z > -0.55)  = 0.0705       (f) P (Z > 3) = 0.0014

(g) P (Z > -3.28) = 0.9995                   (h) P (Z < 4.98) = 0.9999.

Step-by-step explanation:

Let us consider a random variable, [tex]X \sim N (\mu, \sigma^{2})[/tex], then [tex]Z=\frac{X-\mu}{\sigma}[/tex], is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, [tex]Z \sim N (0, 1)[/tex].

In statistics, a standardized score is the number of standard deviations an observation or data point is above the mean.  The z-scores are standardized scores.

The distribution of these z-scores is known as the standard normal distribution.

(a)

Compute the value of P (Z < 2.36) as follows:

P (Z < 2.36) = 0.99086

                   ≈ 0.9909

Thus, the value of P (Z < 2.36) is 0.9909.

(b)

Compute the value of P (Z > 2.36) as follows:

P (Z > 2.36) = 1 - P (Z < 2.36)

                   = 1 - 0.99086

                   = 0.00914

                   ≈ 0.0091

Thus, the value of P (Z > 2.36) is 0.0091.

(c)

Compute the value of P (Z < -1.22) as follows:

P (Z < -1.22) = 0.11123

                   ≈ 0.1112

Thus, the value of P (Z < -1.22) is 0.1112.

(d)

Compute the value of P (1.13 < Z > 3.35) as follows:

P (1.13 < Z > 3.35) = P (Z < 3.35) - P (Z < 1.13)

                            = 0.99960 - 0.87076

                            = 0.12884

                            ≈ 0.1288

Thus, the value of P (1.13 < Z > 3.35)  is 0.1288.

(e)

Compute the value of P (-0.77< Z > -0.55) as follows:

P (-0.77< Z > -0.55) = P (Z < -0.55) - P (Z < -0.77)

                                = 0.29116 - 0.22065

                                = 0.07051

                                ≈ 0.0705

Thus, the value of P (-0.77< Z > -0.55)  is 0.0705.

(f)

Compute the value of P (Z > 3) as follows:

P (Z > 3) = 1 - P (Z < 3)

             = 1 - 0.99865

             = 0.00135

             ≈ 0.0014

Thus, the value of P (Z > 3) is 0.0014.

(g)

Compute the value of P (Z > -3.28) as follows:

P (Z > -3.28) = P (Z < 3.28)

                    = 0.99948

                    ≈ 0.9995

Thus, the value of P (Z > -3.28) is 0.9995.

(h)

Compute the value of P (Z < 4.98) as follows:

P (Z < 4.98) = 0.99999

                   ≈ 0.9999

Thus, the value of P (Z < 4.98) is 0.9999.

**Use the z-table for the probabilities.

To find the probabilities, we use the standard normal distribution table or a calculator to calculate probabilities for a standard normal distribution. We calculate each probability step by step and round the answers to four decimal places.

To determine the probabilities, we will use the standard normal distribution table or a calculator that can calculate probabilities for a standard normal distribution.

(a) P(z < 2.36) = 0.9900

(b) P(z > 2.36) = 1 - P(z < 2.36) = 1 - 0.9900 = 0.0100

(c) P(z < -1.22) = 0.1103

(d) P(1.13 < z < 3.35) = P(z < 3.35) - P(z < 1.13) = 0.9993 - 0.8708 = 0.1285

(e) P(-0.77 < z < -0.55) = P(z < -0.55) - P(z < -0.77) = 0.2896 - 0.2823 = 0.0073

(f) P(z > 3) = 1 - P(z < 3) = 1 - 0.9987 = 0.0013

(g) P(z > -3.28) = 1 - P(z < -3.28) = 1 - 0.0005 = 0.9995

(h) P(z < 4.98) = 1 - P(z > 4.98) = 1 - 0.0000 = 1.0000

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Answer:

$371.39

Step-by-step explanation:

150 * .12+150 = 168

168 * .12+168 = 188.16

188.16 * .12+188.16 = 210.7392

210.7392 * .12+210.7392 = 236.027904

264.3512525 * .12+264.3512525 = 296.0734028

296.0734028 * .12+296.0734028 = 331.6022111

331.6022111 * .12+331.6022111 = 371.3944764

you multiply your current number by 12% and add that to the number, the last number i rounded for the answer as you can see

Answer:

450 cubic meters

Step-by-step explanation:

Austin didn't add the 2 prisms, he multiplied the 1st prism and kept it as the answer.  

Austin is not correct as the combined volume of the prisms is 1100 cubic meter.

Volume is a three-dimensional quantity used to calculate a solid shape's capacity. That means that the volume of a closed form determines how much three-dimensional space it can fill.

From the figure the dimension of prism is 5 x 11 x 10.

Austin says the combined volume of the prisms is 225 cubic meters.

So, Volume of combined prism

= 2 x 5 x 11 x 10

= 2 x 55 x 10

= 110 x 10

= 1100 cubic meter.

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Answer:

a. 20

b. -5

c. 5

Step-by-step explanation:

Please see attachment

Answer:

232 votes

Step-by-step explanation:

This question is just asking what 80% of 290 is. In order to find this, you convert 80% into a decimal, then multiply by 290.

Converting percentages into decimals is just moving the decimal left twice.

0.8 x 290 = 232

You can check this by doing 232/290, and making sure it's 80%.

I hope this helps!

Answer: B The width of the interval would increase

Step-by-step explanation:

i just took this and go the answers back

Decreasing the sample size from 100 to 50 would widen the 95 percent confidence interval for the difference in mean seat widths between two airlines. This is because smaller samples have more variability, requiring a larger interval to capture the population mean with the same level of certainty.

If the consumer group used a sample size of 50 instead of 100 for each airline, that would increase the width of the 95 percent confidence interval for the difference in population mean widths of seats between the two airlines. This is because smaller sample sizes result in more variability, requiring a wider interval to capture the true population mean with the same level of certainty.

As with the unoccupied seats example, where the sample mean of 11.6 and standard deviation of 4.1 were used to form a confidence interval, the size of the interval is dependent on the variability within the sample, and smaller samples generally have higher variability. Similarly in the case of exam scores, with a lower confidence level of 90 percent, a narrower interval is needed compared to a higher confidence level of 95 percent.

Therefore, in effect, decreasing the sample size from 100 to 50 would make the confidence interval wider, as more variability is expected and a larger interval is needed to capture the true population mean with a 95 percent confidence level.

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Step-by-step explanation:

(2575÷25)(1000÷25) = 10340 when reduced to the simplest form. As the numerator is greater than the denominator, we have an IMPROPER fraction, so we can also express it as a MIXED NUMBER, thus 25751000 is also equal to 22340 when expressed as a mixed number.

Answer:

[tex]s^2=1.32[/tex]

Step-by-step explanation:

-This is a binomial probability distribution problem.

-Given that p=0.86 and n=11, the sample variance can be calculated using the formula:

[tex]\sigma^2=np(1-p)[/tex]

#We substitute the given parameters in the formula to solve for variance:

[tex]\sigma^2=np(1-p)\\\\\\=11\times 0.86(1-0.86)\\\\=1.324\approx1.32[/tex]

Hence, the sample variance is 1.32

Final answer:

The variance in a sample of 11 flights that have arrived on time at Denver International Airport, where 86% of flights are on time, is calculated using a binomial distribution with the formula variance = np(1-p), resulting in a variance of 1.33 after rounding to two decimal places.

Explanation:

The subject of this question is about finding the variance in a sample of flights that have arrived on time at Denver International Airport, given that 86% of recent flights have arrived on time and that a sample of 11 flights is studied. To calculate the variance for a binomial distribution, which is applicable in this context because each flight can either be on time or not, we use the formula:

Variance = np(1-p).

Where 'n' is the number of trials (or flights, in this case), which is 11, and 'p' is the probability of success on each trial (a flight arriving on time), which is 0.86. Thus, the variance for the 11 flights can be calculated as follows:

Variance = 11 × 0.86 × (1 - 0.86) = 11 × 0.86 × 0.14 = 1.3284.

After rounding to two decimal places, the variance for a sample of 11 flights is 1.33.

Answer: don’t use k for the other numbers

Step-by-step explanation:

Answer:

6k -5

Step-by-step explanation:

19-6(-k+4)

Distribute

19 -6*-k -6*4

19 +6k -24

Combine like terms

6k +19-24

6k -5

Answer:

Find the difference between the two scores for a number of sample distributions. Make a plot of the differences and check for outliers.

Step-by-step explanation:

Checking for Normality means basically checking if one's data distribution approximates a normal distribution.

A normal distribution is represented by a bell-shaped curve, peaking around the mean, indicating that all of the data spreads out from the mean.

Th original aim of the teacher is to check the effects of the particular added unit on the performance of students in the subject.

The teacher goes about this by testing the students before and after learning the unit.

The best way to compare of course, is to take a difference of the test scores for different samples. This first gives the idea of whether the newly introduced unit affects performance.

This set of differences is then checked for normality.

So, the best manner to make a plot of these differences. Like we mentioned earlier, a normal distribution is bell shaped. So, the plot of these differences would be a bell shaped curve if the distribution was normal and we wouldn't get a bell shaped curve if the distribution wasn't normal.

Checking for outliers help to eliminate part of data that can totally scatter the regular behaviour of the data distribution.

So, the best way for the teacher to check for normality is to find the difference between the two scores for a number of sample distributions. Make a plot of the differences and check for outliers.

Hope this Helps!!!

Since the data include the pretest and post test scores of each student or subject, then, it is approached as a paired t test. Hence, checking for normality would require obtaining the difference between the two scores for each subject, then make a plot of the differences and check for outliers.

Hence, the most appropriate check for normality in this scenario is to check for outliers in the plot made from the score difference.

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Given:

Given that Brinson's family is going camping. Their tent is shaped like a rectangular pyramid.

The volume of the tent is 6000 cubic inches.

The area of the base of the tent is 1200 square inches.

We need to determine the height of the tent.

Height of the tent:

The height of the tent can be determined using the formula,

[tex]V=\frac{1}{3}Bh[/tex]

where B is the area of the base and h is the height of the pyramid.

Substituting V = 6000 and B = 1200, we get;

[tex]6000=\frac{1}{3}(1200)h[/tex]

[tex]6000=400h[/tex]

   [tex]15=h[/tex]

Thus, the height of the tent is 15 inches.

Answer:

Step-by-step explanation:

Final answer:

The value of sin(3π/2) is -1, and the value of cos(3π/2) is 0, as determined by their positions on the unit circle at an angle of 3π/2 radians.

Explanation:

To find the values of sin(3π/2) and cos(3π/2) using the unit circle, we need to understand the positions on the circle.

At 3π/2 radians, which is the same as 270°, the point on the unit circle is directly below the center of the circle, at coordinates (0, -1).

This position corresponds to a cosine value of 0 (since the x-coordinate is 0) and a sine value of -1 (since the y-coordinate is -1).

Answer: x = -6

Step-by-step explanation: answer on edge

Answer:

1. y = f(8)

2. So for t which f'(t) > 0, the drug level in the bloodstream is increasing. And for t which f'(t) < 0, it is decreasing.

Step-by-step explanation:

The concentration of the drug in the bloodstream at t hours is:

y = f(t)

1. What is the concentration of the drug in the bloodstream at t= 8 hours?

At t hours, y = f(t)

So at 8 hours, y = f(8)

2. During what time interval is the drug level in the bloodstream increasing? Decreasing?

A function f(t) is increasing when

f'(t) > 0

And is decreasing when

f'(t) < 0

So for t which f'(t) > 0, the drug level in the bloodstream is increasing. And for t which f'(t) < 0, it is decreasing.

(1) The concentration  of drug in the bloodstream in 8 hours is given by

[tex]\rm \bold{y = f (8)}[/tex]

(2) The time interval for which [tex]\rm y'=f'(t)>0[/tex] the drug level of the bloodstream is increasing.

The time interval for which [tex]\rm y' = f'(t) <0[/tex] the drug level of the bloodstream is decreasing.

When a certain prescription drug is taken orally by an adult.

the amount of the drug (in mg/L) in the bloodstream at t hours is given by the function y=f(t)

To be determined

(1) The concentration of the drug in the bloodstream at t= 8 hours

(2) During what time interval is the drug level in the bloodstream increasing or deceasing

The amount of the drug (in mg/L) in the bloodstream at t hours is given by the function

y=f(t).......(1)

(1)  The concentration  of drug in the bloodstream in 8 hours is given by putting t= 8 in the equation (1) which can be formulated as below

[tex]\rm y = f (8)[/tex]

(2) From the definition of increasing and decreasing function we can write that

[tex]\rm y = f(x) \; is \; increasing \; when \; f' (x)>0 \\and\; y = f(x) ; is \; decreasing \; when \; f' (x)<0 \\\\\\\\[/tex]

By the definition of increasing and decreasing function we can say that

The time interval for which [tex]\rm y'=f'(t)>0[/tex] the drug level of the bloodstream is increasing.

The time interval for which [tex]\rm y'=f'(t) <0[/tex] the drug level of the blood stream is decreasing.

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Rounding to the nearest minute, your visit to the doctor will take approximately 36 minutes.

To determine how long your visit to the doctor will take, we need to find the time it takes for the remaining amount of radioactive dye in your system to be less than 2 milligrams.

We can use the exponential decay formula to model the amount of radioactive dye remaining in your system after a given time:

[tex]\[ A(t) = A_0 \times e^{-kt} \][/tex]

Where:

- [tex]\( A(t) \)[/tex] is the amount of radioactive dye remaining at time ( t )

- [tex]\( A_0 \)[/tex] is the initial amount of radioactive dye

- [tex]\( k \)[/tex] is the decay constant

- [tex]\( t \)[/tex] is the time (in minutes, in this case)

Given:

- [tex]\( A_0 = 11 \)[/tex] milligrams

- [tex]\( A(t) = 4.25 \)[/tex] milligrams

-[tex]\( t = 20 \)[/tex] minutes

We need to solve for ( k ) using the given data:

[tex]\[ 4.25 = 11 \times e^{-20k} \][/tex]

First, divide both sides by 11:

[tex]\[ \frac{4.25}{11} = e^{-20k} \][/tex]

Now, take the natural logarithm of both sides:

[tex]\[ \ln\left(\frac{4.25}{11}\right) = -20k \][/tex]

Now, solve for [tex]\( k \):[/tex]

[tex]\[ k = \frac{\ln\left(\frac{4.25}{11}\right)}{-20} \][/tex]

Now that we have the decay constant, we can use it to find the time it takes for the remaining amount of dye to be less than 2 milligrams:

[tex]\[ 2 = 11 \times e^{-kt} \][/tex]

Plug in the value of ( k ) and solve for ( t ):

[tex]\[ t = \frac{\ln\left(\frac{2}{11}\right)}{-k} \][/tex]

Let's calculate ( t ):

First, let's calculate ( k ):

[tex]\[ k = \frac{\ln\left(\frac{4.25}{11}\right)}{-20} \][/tex]

[tex]\[ k \approx \frac{\ln(0.3864)}{-20} \][/tex]

[tex]\[ k \approx \frac{-0.9515}{-20} \][/tex]

[tex]\[ k \approx 0.0476 \][/tex]

Now, let's use ( k ) to find ( t ):

[tex]\[ t = \frac{\ln\left(\frac{2}{11}\right)}{-0.0476} \][/tex]

[tex]\[ t \approx \frac{\ln(0.1818)}{-0.0476} \][/tex]

[tex]\[ t \approx \frac{-1.707}{-0.0476} \][/tex]

[tex]\[ t \approx 35.81 \][/tex]

Rounding to the nearest minute, your visit to the doctor will take approximately 36 minutes.

Answer: I had a stroke trying to read and understand this sorry

Step-by-step explanation:

Answer:

$60.

Step-by-step explanation:

x / 2 - 10 = 0

x / 2 = 10

x = 20

Before she spent her money:

x / 2 - 10 = 20

x / 2 = 30

x = 60

She had $60 before she went shopping.

Feel free to let me know if you need more help. :)

Answer:

Cube surface area = 6 * side^2

150 square inches = 6 * side^2

sq root (side) = 25

side = 5 inches

Step-by-step explanation:

Answer:

g(-1) = -3

Step-by-step explanation:

g(-1) is the y value of the graph when x = -1

when x=-1 y=-3

g(-1) = -3

Answer:

[tex]15.55-2.997\frac{0.278}{\sqrt{8}}=15.26[/tex]    

[tex]15.55+2.997\frac{0.278}{\sqrt{8}}=15.84[/tex]    

The 98% confidence interval would be given by (15.26;15.84)    

Step-by-step explanation:

Notation

[tex]\bar X[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

We can calculate the mean and the sample deviation we can use the following formulas:  

[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)  

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)  

The mean calculated for this case is [tex]\bar X=15.55[/tex]

The sample deviation calculated [tex]s=0.278[/tex]

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=8-1=7[/tex]

Since the Confidence is 0.98 or 98%, the value of [tex]\alpha=0.02[/tex] and [tex]\alpha/2 =0.01[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,7)".And we see that [tex]t_{\alpha/2}=2.997[/tex]

And the confidence interval is given by:

[tex]15.55-2.997\frac{0.278}{\sqrt{8}}=15.26[/tex]    

[tex]15.55+2.997\frac{0.278}{\sqrt{8}}=15.84[/tex]    

The 98% confidence interval would be given by (15.26;15.84)    

Answer:

z = 1.960

Step-by-step explanation:

The sample proportion is:

p = 715 / 2684 = 0.2664

The standard error is:

σ = √(pq/n)

σ = √(0.266 × 0.734 / 2684)

σ = 0.0085

For α = 0.05, the confidence level is 95%.  The z-statistic at 95% confidence is 1.960.

The margin of error is 1.960 × 0.0085 = 0.0167.

The confidence interval is 0.2664 ± 0.0167 = (0.2497, 0.2831).

The upper limit is 28.3%, so the journal can conclude with 95% confidence that the true percentage is less than 29%.

Yes, the considered sample provides sufficient evidence to indicate that the true percentage of all firms that announced one or more acquisitions during the year 2000 is less than 29​%.

Suppose that we have:

Then, the z test statistic for one sample proportion is:

[tex]Z = \dfrac{\hat{p} - p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]

For this case, we're provided that:

We want to determine if true percentage of all firms that announced one or more acquisitions during the year 2000 is less than 29​% = 0.29 (converted percent to decimal).

Hypotheses:

Thus, the test is left tailed test.

where [tex]p_0[/tex] = 29% = 0.29 is the hypothesized mean value of population proportion.

The test statistic is:

[tex]Z = \dfrac{\hat{p} - p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\\\ Z = \dfrac{715/2684 - 0.29}{\sqrt{\dfrac{0.29(1-0.29)}{2684}}} \approx -2.69[/tex]

The critical value of Z at level of significance 0.05 is -1.6449

Since the test statistic = -2.69 < critical value = -1.6449, so the test statistic lies in the rejection region (the rejection region for the left tailed test is all the values below critical value).

Thus, we reject the null hypothesis and accept the alternative hypothesis that the true population mean is less than 0.29.

Thus, the considered sample provides sufficient evidence to indicate that the true percentage of all firms that announced one or more acquisitions during the year 2000 is less than 29​%.

The z-test statistic came out to be -2.69

Learn more about one-sample z-test for population proportion mean here:

https://brainly.com/question/16173147

Answer:

[tex]x+3y=-27[/tex]

Step-by-step explanation:

We are given that

[tex]y=(-\frac{1}{3})x-9[/tex]

We have to find the  standard form of given equation

[tex]y=\frac{-x-27}{3}[/tex]

[tex]3y=-x-27[/tex]

By using multiplication property of equality

[tex]x+3y=-27[/tex]

We know that

Standard form of equation

[tex]ax+by=c[/tex]

Therefore, the standard form of given equation  is given by

[tex]x+3y=-27[/tex]

The equation y = -1/3x - 9 can be converted to standard form by eliminating fractions and rearranging. The final equation in standard form is x - 3y = 27.

The equation you provided is already in slope-intercept form, which is y = mx + b, where m is the slope and b is the y-intercept. However, you're asked to convert this equation into standard form, which is Ax + By = C, with A, B, and C being integers, and A and B not both equal to zero.

To convert the given equation y = -1/3x - 9 into standard form, we first need to eliminate the fractions. We can achieve this by multiplying every term by -3, giving us 3y = x + 27. To make it fit the standard form, we can rearrange as -x + 3y = -27.

Remember, standard form shouldn't have any negatives in front of the x term, so we multiply everything by -1. The final equation in standard form is x - 3y = 27.

https://brainly.com/question/16646502

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Answer:

(a) 95% confidence interval for the percentage of all car accidents that involve teenage drivers is [0.177 , 0.243].

(b) We are 95% confident that the percentage of all car accidents that involve teenage drivers will lie between 17.7% and 24.3%.

(c) We conclude that the the percentage of teenagers has not changed since you join the company.

(d) We conclude that the the percentage of teenagers has changed since you join the company.

Step-by-step explanation:

We are given that your manager asked you to check police records of car accidents and out of 576 accidents you selected randomly, teenagers were at the wheel in 120 of them.

(a) Firstly, the pivotal quantity for 95% confidence interval for the population proportion is given by;

                        P.Q. = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} }}[/tex]  ~ N(0,1)

where, [tex]\hat p[/tex] = sample proportion teenage drivers = [tex]\frac{120}{576}[/tex] = 0.21

           n = sample of accidents = 576

           p = population percentage of all car accidents

Here for constructing 95% confidence interval we have used One-sample z proportion statistics.

So, 95% confidence interval for the population population, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                     of significance are -1.96 & 1.96}  

P(-1.96 < [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} }}[/tex] < 1.96) = 0.95

P( [tex]-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}[/tex] < [tex]{\hat p-p}[/tex] < [tex]1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}[/tex] ) = 0.95

P( [tex]\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}[/tex] < p < [tex]\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}[/tex] ) = 0.95

95% confidence interval for p = [[tex]\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}[/tex] , [tex]\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}[/tex]]

  = [ [tex]0.21-1.96 \times {\sqrt{\frac{0.21(1-0.21)}{576} }}[/tex] , [tex]0.21+1.96 \times {\sqrt{\frac{0.21(1-0.21)}{576} }}[/tex] ]

  = [0.177 , 0.243]

Therefore, 95% confidence interval for the percentage of all car accidents that involve teenage drivers is [0.177 , 0.243].

(b) We are 95% confident that the percentage of all car accidents that involve teenage drivers will lie between 17.7% and 24.3%.

(c) We are also provided that before you were hired in the company, the percentage of teenagers who where involved in car accidents was 18%.

The manager wants to see if the percentage of teenagers has changed since you join the company.

Let p = percentage of teenagers who where involved in car accidents

So, Null Hypothesis, [tex]H_0[/tex] : p = 18%    {means that the percentage of teenagers has not changed since you join the company}

Alternate Hypothesis, [tex]H_A[/tex] : p [tex]\neq[/tex] 18%    {means that the percentage of teenagers has changed since you join the company}

The test statistics that will be used here is One-sample z proportion statistics;

                              T.S.  =  [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} }}[/tex]  ~ N(0,1)

where, [tex]\hat p[/tex] = sample proportion teenage drivers = [tex]\frac{120}{576}[/tex] = 0.21

           n = sample of accidents = 576

So, test statistics  =  [tex]\frac{0.21-0.18}{\sqrt{\frac{0.21(1-0.21)}{576} }}[/tex]  

                              =  1.768

The value of the sample test statistics is 1.768.

Now at 0.05 significance level, the z table gives critical value of -1.96 and 1.96 for two-tailed test. Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the the percentage of teenagers has not changed since you join the company.

(d) Now at 0.1 significance level, the z table gives critical value of -1.6449 and 1.6449 for two-tailed test. Since our test statistics does not lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the the percentage of teenagers has changed since you join the company.