<h2>
Answer with explanation:</h2>
Let p be the population proportion of parents who had children in grades K-12 were satisfied with the quality of education the students receive.
Given : Several years ago, 39% of parents who had children in grades K-12 were satisfied with the quality of education the students receive.
Set hypothesis to test :

Sample size : n= 1055
Sample proportion : 
Critical value for 95% confidence : 
Confidence interval : 

Since , Confidence interval does not contain 0.39.
It means we reject the null hypothesis.
We conclude that 95% confidence interval represents evidence that parents' attitudes toward the quality of education have changed.
If you plug that into a calculator you get 4.5
Answer:
The probability that a randomly selected chocolate bar will have a. Between 200 and 220 calories will be 0.3023
Step-by-step explanation:
Given:
True mean=225
S.D=10
X1=200 and X2=220
To Find:
P(X1<x<X2)
Solution:
<em>BY using Z-table for probability and Z-score we can proceed.</em>
So
For 200 calories Z will be ,
Z=(sample mean -true mean)/S.D
=200-225/10
=-25/10
=-2.5
For 220 calories Z will be ,
Z=220-225/10
=-5/10
=-0.5
So required probability will be




=0.3023
Step-by-step explanation:
Given
g(x) = - 2x² + 3x
g(3) = - 2 * 3² + 3 *3
= -2 * 9 + 9
= -18 + 9
= -9