How to multiply to make an equivalent fraction

Please help, I’ll mark you as brainliest!!!!!

kolezko [41]

Answer:10

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

A study has shown that the daily amount of milk produced by a dairy cow is

8090 [49]

Answer:

A

Step-by-step explanation:

this is more of a mind game. Lots of words to throw you off. Daily a cow produces a mean of 6.2 gallons (average of 6.2 gallons, meaning they already found the mean, "averaged" it out for you) with a deviation of .7 gallons ("give or take" .7 gallons). On any given day (throw these words out) 68% of the cows (don't need this information to answer the question either, thrown in to confuse you) will produce an amount of milk within which of the following ranges? Just subtract the deviation ("give or take number") .7 gallons from the mean (average (already determined)) 6.2 gallons which is 5.5 gallons THEN add the deviation ("give or take number") .7 gallons to the mean (average (already determined)) 6.2 gallons which is 6.9 gallons. Your answer is A 5.5 gallons to 6.9 gallons. You don't even have to go crazy on the math on this question you can rule them all out but A immediately with subtracting .7 from 6.2. Mind games .... LOL sneaky.

6 0

3 years ago

During the first 3 days of the week, Mike read an average of 28 pages per day. During the next 4 days, Mike averaged 42 pages pe

Anestetic [448]

Answer:

36 pages per day on average

Step-by-step explanation:

Calculate the average.

3 * 28 (The first 3 days of the week where Mike reads 28 pages per day) = 84

4 * 42 (The next 4 days where Mike reads 42 pages per day ) = 168

168 + 84 = 252

To find the average using this total, divide it by the total amount of days of reading (seven days in this problem).

252 / 7 = 36

6 0

2 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

Brainiest for whoever helps on both questions first in next ten mins

Nana76 [90]

1. D

2. C

3. D

4. A

5. C

6. D

7. B

8. C

9. A

10. B

100% correct just took the quiz!!!

6 0

3 years ago