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krok68 [10]
3 years ago
9

Please help will mark brainliest if correct

Mathematics
2 answers:
balandron [24]3 years ago
8 0
The answer to your question is D
Ipatiy [6.2K]3 years ago
3 0

Answer:

D.

Step-by-step explanation:

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Derek reads at a rate of 25 words per minute. If he reads constantly until he is finished how long would it take him to read a 2
sweet-ann [11.9K]

Answer:

9

Step-by-step explanation:

if you divide 225 by 25 u get 9

5 0
3 years ago
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
Of all the rectangles with a perimeter of 168 feet, find the dimension of the one with the largest area
dolphi86 [110]
<span>The rectangle with the largest area with a given perimeter will be a square - so the sides will be equal. So we need to find length of side, L, such that 4*L=168. L = 168/4 L=42. So the dimensions of the rectangle that maximizes the area with a perimiter of 168 feet are: 42 feet by 24 feet.</span>
3 0
4 years ago
F(x)=-3/5x+k and if f(30)=-11 what is the value of k
Kruka [31]

Answer:

k=20.

Explanation: First we find where f(x) has its local extrema: f'(x)=3x2−10x+3. The critical points are roots of the equation: 3x2−10x+3=0.

6 0
3 years ago
Write a unit rate for the situation.<br> 6 kittens in 3 boxes
MrRa [10]

Answer:

6:3

Step-by-step explanation:

7 0
4 years ago
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