How do you write 9.01 x 103 in standard form?

Because of safety considerations, in May 2003 the Federal Aviation Administration (FAA) changed its guidelines for how small com

Rasek [7]

Answer:

a) The 95% confidence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers is between 179 and 187 pounds. This means that we are 95% sure that the mean summer weight of all Frontier Airlines passengers is between these two values.

b)

The 95% confidence interval for the mean winter weight (including carry-on luggage) of Frontier Airlines passengers is between 185.4 pounds and 194.6 pounds. This means that we are 95% sure that the mean winter weight of all Frontier Airlines passengers is between these two values.

c) They are respected, as the upper bound of both intervals is below the new FAA recommendations.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve these questions.

Question a:

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 100 - 1 = 99

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 99 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$ . So we have T = 1.9842

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.9842\frac{20}{\sqrt{100}} = 4$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 183 - 4 = 179 pounds.

The upper end of the interval is the sample mean added to M. So it is 183 + 4 = 187 pounds.

The 95% confidence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers is between 179 and 187 pounds. This means that we are 95% sure that the mean summer weight of all Frontier Airlines passengers is between these two values.

Question b:

Critical value is the same(same sample size and confidence level).

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.9842\frac{23}{\sqrt{100}} = 4.6$

The lower end of the interval is the sample mean subtracted by M. So it is 190 - 4.6 = 185.4 pounds.

The upper end of the interval is the sample mean added to M. So it is 190 + 4.6 = 194.6 pounds.

The 95% confidence interval for the mean winter weight (including carry-on luggage) of Frontier Airlines passengers is between 185.4 pounds and 194.6 pounds. This means that we are 95% sure that the mean winter weight of all Frontier Airlines passengers is between these two values.

c. The new FAA recommendations are 190 pounds for summer and 195 pounds for winter. Comment on these recommendations in light of the confidence interval estimates from Parts (a) and (b).

They are respected, as the upper bound of both intervals is below the new FAA recommendations.

7 0

3 years ago

which property would allow you to use mental computation to simply the problem the problem 27 +15 +3 +5

PilotLPTM [1.2K]

Answer:

Addition

Step-by-step explanation:

4 0

3 years ago

If the unicorn seat on a merry–go–round is 5 feet from the center, and the merry–go–round completes a full rotation in 6 seconds

Cerrena [4.2K]

The answer is 5.23 ft/second

The way I got this answer was first finding the circumference of the merry-go-round using the equation $C=2 \pi r$ , where r represents the radius, which is the distance from the center of a circle.

So your equation should look like this $C=2 \pi(5)$ . I used the rounded term of 3.14 for $\pi$

Once you multiply those terms, you should get the answer, 31.4. Now this is the total distance around the merry-go-round in feet. So, if it takes 6 seconds to make a full rotation and we are trying to find the amount offeet traveled in one second, we should divide the circumference (31.4) by 6 to give us our answer.

So, 31.4÷6=5.2333...ft/second

And I rounded the answer to the nearest hundredth to get 5.23 feet/second

6 0

3 years ago

Read 2 more answers

f the probability of being hospitalized during a year is 0.25 , find the probability that no one in a family of five will be ho

Kaylis [27]

Answer:

The probability that no one in a family of five will be hospitalized in a year is 0,2373

Step-by-step explanation:

Let p denote the probability of being hospitalized during a year,

Then p=0.25

And p' the probability of not being hospitalized during a year is 1-p

Thus p' = 1-p = 1-0.25 = 0.75

The probability that no one in a family of five will be hospitalized in a year is equal to $p'^{5}$ . That is

$p'^{5}$ = $0.75^{5}$ ≈ 0,2373

3 0

3 years ago

Which of the following is the logical conclusion to the conditional statements below? a arrow b mb arrow c

IRINA_888 [86]

Answer:

B

Step-by-step explanation:

From what we have here;

if a implies b and b implies c

It simply means that we have a implying c

the correct representation here is given in the second option

And thus, we have it that;

a => c

5 0

4 years ago