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alexandr1967 [171]

2 years ago

7

One week, Kerry travels 125 miles and uses 5 gallons of gasoline. The next week, she travels 175 miles and uses 7 gallons of gas

oline.

1 answer:

bazaltina [42]2 years ago

6 0

This is good information, but the question part is missing. Put the question in the comments and I'll try my best to help!

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What is the value of y=4x−2<br> when x=3<br> ?

MariettaO [177]

Answer:

y = 10

Step-by-step explanation:

y = 4(3) - 2

y = 12 - 2

y = 10

7 0

3 years ago

Read 2 more answers

Hans is on his way home in his car. He has driven 8 miles so far, which is one-fourth of the way home. What is the total length

icang [17]

8x4= 32miles

Hope you are satisfied

6 0

2 years ago

Which of the following is equivalent to “the product of 2 and 3 is increased by 7”?

trasher [3.6K]

A. 2(3)+7

product is multiplication
increased is addition

8 0

2 years ago

A tennis player gets an ace on 35% of his serves. Out of 80 serves about how many aces will he get

kari74 [83]

Answer:

28

Step-by-step explanation:

His aces on 80 serve = 35% of 80

= 35/100 × 80

= 28

7 0

2 years ago

Read 2 more answers

In 2002, the mean age of an inmate on death row was 40.7 years with a standard deviation of 9.6 years according to the U.S. Depa

marissa [1.9K]

Answer:

The <em>95% confidence interval</em> for the current mean age of death-row inmates is between 42.23 years and 35.57 years.

Step-by-step explanation:

The <em>confidence interval</em> of the mean is given by the next formula:

$\\ \overline{x} \pm z_{1-\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$ [1]

We already know (according to the U.S. Department of Justice):

The (population) standard deviation for this case (mean age of an inmate on death row) has a standard deviation of 9.6 years ( $\\ \sigma = 9.6$ years).
The number of observations for the sample taken is $\\ n = 32$ .
The sample mean, $\\ \overline{x} = 38.9$ years.

For $\\ z_{1-\frac{\alpha}{2}}$ , we have that $\\ \alpha = 0.05$ . That is, the <em>level of significance</em> $\\ \alpha$ is 1 - 0.95 = 0.05. In this case, then, we have that the <em>z-score</em> corresponding to this case is:

$\\ z_{1-\frac{\alpha}{2}} = z_{1-\frac{0.05}{2}} = z_{1-0.025} = z_{0.975}$

Consulting a cumulative <em>standard normal table</em>, available on the Internet or in Statistics books, to find the z-score associated to the probability of, $\\ P(z$ , we have that $\\ z = 1.96$ .

Notice that we supposed that the sample is from a population that follows a <em>normal distribution</em>. However, we also have a value for n > 30, and we already know that for this result the sampling distribution for the sample means follows, approximately, a normal distribution with mean, $\\ \mu$ , and standard deviation, $\\ \sigma_{\overline{x}} = \frac{\sigma}{\sqrt{n}}$ .

Having all this information, we can proceed to answer the question.

Constructing the 95% confidence interval for the current mean age of death-row inmates

To construct the 95% confidence interval, we already know that this interval is given by [1]:

$\\ \overline{x} \pm z_{1-\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$

That is, we have:

$\\ \overline{x} = 38.9$ years.

$\\ z_{1-\frac{\alpha}{2}} = 1.96$

$\\ \sigma = 9.6$ years.

$\\ n = 32$

Then

$\\ 38.9 \pm 1.96*\frac{9.6}{\sqrt{32}}$

$\\ 38.9 \pm 1.96*\frac{9.6}{5.656854}$

$\\ 38.9 \pm 1.96*1.697056$

$\\ 38.9 \pm 3.326229$

Therefore, the Upper and Lower limits of the interval are:

Upper limit:

$\\ 38.9 + 3.326229$

$\\ 42.226229 \approx 42.23$ years.

Lower limit:

$\\ 38.9 - 3.326229$

$\\ 35.573771 \approx 35.57$ years.

In sum, the 95% confidence interval for the current mean age of death-row inmates is between 42.23 years and 35.57 years.

Notice that the "mean age of an inmate on death row was 40.7 years in 2002", and this value is between the limits of the 95% confidence interval obtained. So, according to the random sample under study, it seems that this mean age has not changed.

7 0

2 years ago