Given:
Kelly purchased 5 containers of ice cream for a party.
Each container holds 8 cups of ice cream.
Her brother ate 2 cups of ice cream.
One serving of ice cream = 2/3 of a cup.
To find:
Number of servings of ice cream will be left for the party?
Solution:
We have,
1 container = 8 cups of ice cream
5 containers = 5×8 = 40 cups of ice cream
Her brother ate 2 cups of ice cream. So, remaining cups of ice cream is
cups
Now,
of a cup = 1 serving
1 cup =
serving
38 cup =
serving
=
serving
= 57 serving
Therefore, 57 servings of ice cream will be left for the party.
Using the z-distribution, it is found that the 95% confidence interval to estimate the mean SAT math score in this state for this year is (472, 488).
We have the <u>standard deviation for the population</u>, which is why the z-distribution is used to solve this question.
- The sample mean is
. - The population standard deviation is
. - The sample size is
.
The interval is given by:
![\overline{x} \pm z\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%20%5Cpm%20z%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
We have to find the critical value, which is z with a p-value of
, in which
is the confidence level.
In this problem,
, thus, z with a p-value of
, which means that it is z = 1.96.
Then:
![\overline{x} - z\frac{\sigma}{\sqrt{n}} = 480 - 1.96\frac{100}{\sqrt{656}} = 472](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%20-%20z%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%20%3D%20480%20-%201.96%5Cfrac%7B100%7D%7B%5Csqrt%7B656%7D%7D%20%3D%20472)
![\overline{x} + z\frac{\sigma}{\sqrt{n}} = 480 + 1.96\frac{100}{\sqrt{656}} = 488](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%20%2B%20z%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%20%3D%20480%20%2B%201.96%5Cfrac%7B100%7D%7B%5Csqrt%7B656%7D%7D%20%3D%20488)
The 95% confidence interval to estimate the mean SAT math score in this state for this year is (472, 488).
A similar problem is given at brainly.com/question/22596713
Answer:
-i
Step-by-step explanation:
i^0=1
i^1=i
i^2=-1
i^3=-i
i^4=1
This repeats so we want to see how many 4 factors of i there is in i^(23) which is 5 with a remainder of 3.
So i^(23)=i^3=-i.
i^(23)=i^(5*4+3)=(i^4)^5 * (i^3)=(1)^5 * (-i)=1(-i)=-i.