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3 years ago

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At a large automobile dealership, you take a random sample of 35 cars and determine that the average cost is $34,330. Which stat

ement is true?

1 answer:

nadya68 [22]3 years ago

4 0

Answer:Btw id your on Usa Testprep the answer is d

Step-by-step explanation:

You might be interested in

The sum of the two angles of a triangle is 80 degree and their difference is 20 degree. find all the angles of the triangle.

Norma-Jean [14]

Sum of 2 angles = 80°

Difference = 20°

If we take away 20° from the sum, both angles are equal

⇒80 - 20 = 60°

Divide by 2 to find the smaller angle:

60 ÷ 2 = 30

Add 20° to find the bigger angle:

20 + 30 = 50°

<u>One of the angle is 30° and the other angle is 50°</u>

Given that the sum of the two angles is 80°

⇒ Third angle = 180 - 80 = 100°

Answer: The 3 angles are 30° , 50° and 100°

3 0

3 years ago

Read 2 more answers

One person wants to get a 95% z-confidence interval with a margin of error of at most 15 based on a population standard deviatio

Bess [88]

Answer:

$n=(\frac{1.96(60)}{15})^2 =61.46 \approx 62$

So the answer for this case would be n=62 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean

$\sigma=60$ represent the population standard deviation

n represent the sample size (variable of interest)

Confidence =95% or 0.95

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

And on this case we have that ME =15, and we are interested in order to find the value of n, if we solve n from equation (1) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (2)

The critical value for 95% of confidence interval is provided, $z_{\alpha/2}=1.96$ , replacing into formula (2) we got:

$n=(\frac{1.96(60)}{15})^2 =61.46 \approx 62$

So the answer for this case would be n=62 rounded up to the nearest integer

6 0

3 years ago

Please help, the problem is bellow

Anuta_ua [19.1K]

4x + 1.25y = 500. Hope this helps

6 0

2 years ago

What is 4x • 6y =9c.

LuckyWell [14K]

C=8/3 xy

Downloaded Photomath. That app is the best. Shows you how to get the answer along with the answer so you can learn how to do it for test.

6 0

2 years ago

You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confide

salantis [7]

Answer:

With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the 90%.

Step-by-step explanation:

We are given that a random sample of 60 home theater systems has a mean price of$131.00. Assume the population standard deviation is$18.80.

Firstly, the pivotal quantity for 90% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean price = $131

$\sigma$ = population standard deviation = $18.80

n = sample of home theater = 60

$\mu$ = population mean

<em>Here for constructing 90% confidence interval we have used One-sample z test statistics as we know about the population standard deviation.</em>

<u>So, 90% confidence interval for the population mean, </u> $\mu$ <u> is ;</u>

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

<u>90% confidence interval for</u> $\mu$ = [ $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $131-1.645 \times {\frac{18.8}{\sqrt{60} } }$ , $131+1.645 \times {\frac{18.8}{\sqrt{60} } }$ ]

= [127.01 , 134.99]

Therefore, 90% confidence interval for the population mean is [127.01 , 134.99].

Now, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean price = $131

$\sigma$ = population standard deviation = $18.80

n = sample of home theater = 60

$\mu$ = population mean

<em>Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about the population standard deviation.</em>

<u>So, 95% confidence interval for the population mean, </u> $\mu$ <u> is ;</u>

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 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

<u>95% confidence interval for</u> $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $131-1.96 \times {\frac{18.8}{\sqrt{60} } }$ , $131+1.96 \times {\frac{18.8}{\sqrt{60} } }$ ]

= [126.24 , 135.76]

Therefore, 95% confidence interval for the population mean is [126.24 , 135.76].

Now, with 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the 90%.

7 0

3 years ago