38/244= about 0.16
24/943= about 0.03
653/53= about 12.32
All answers are rounded
Answer:
24 ounces of butter
21 ounces of sugar
30 ounces of flour
12 eggs
3 teaspoon of baking powder
Step-by-step explanation:
Simple. All you had to do was multiply each quantity by 3, because you needed 3 lots of each.
Hope this helps.
Answer:
The 95% confidence interval for the true average number of homes that a person owns in his or her lifetime is (4,6.2).
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom,which is the sample size subtracted by 1. So
df = 50 - 1 = 49
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 49 degrees of freedom(y-axis) and a confidence level of
. So we have T = 2.0096
The margin of error is:
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 5.1 - 1.1 = 4
The upper end of the interval is the sample mean added to M. So it is 5.1 + 1.1 = 6.2.
The 95% confidence interval for the true average number of homes that a person owns in his or her lifetime is (4,6.2).
By Stokes' theorem,

where

is the circular boundary of the hemisphere

in the

-

plane. We can parameterize the boundary via the "standard" choice of polar coordinates, setting

where

. Then the line integral is


We can check this result by evaluating the equivalent surface integral. We have

and we can parameterize

by

so that

where

and

. Then,

as expected.
She was at 40 feet below, she swam up getting so she was then at 40-10 = 30 feet below sea level.
She the swan 4 feet down so she ended up at 30 + 4 = 34 feet below sea level.
The answer is 34 feet below sea level (-34)