2(5b-7), you can input b as 3y
2(5(3y)-7) —> 2(15y-7) —> 30y -14
—> 30y = 14 —> y = 14/30 —> y = 4.6667
first, lets distribute the equation. 5(n^2+n)-3n(2n^2+4n-2) would become
5n^2 +5n -6n^3 -12n^2 +6n
Now we add the like terms.
-6n^3 - 7n^2 +11n
You have given us no information whatsoever concerning the distribution
of M&Ms in Matthew's pencil box, so it's quite impossible to formulate any
sort of answer to the question.
Answer:
C. H0 : p = 0.8 H 1 : p ≠ 0.8
The test is:_____.
c. two-tailed
The test statistic is:______p ± z (base alpha by 2) 
The p-value is:_____. 0.09887
Based on this we:_____.
B. Reject the null hypothesis.
Step-by-step explanation:
We formulate null and alternative hypotheses as proportion of people who own cats is significantly different than 80%.
H0 : p = 0.8 H 1 : p ≠ 0.8
The alternative hypothesis H1 is that the 80% of the proportion is different and null hypothesis is , it is same.
For a two tailed test for significance level = 0.2 we have critical value ± 1.28.
We have alpha equal to 0.2 for a two tailed test . We divided alpha with 2 to get the answer for a two tailed test. When divided by two it gives 0.1 and the corresponding value is ± 1.28
The test statistic is
p ± z (base alpha by 2) 
Where p = 0.8 , q = 1-p= 1-0.8= 0.2
n= 200
Putting the values
0.8 ± 1.28 
0.8 ± 0.03620
0.8362, 0.7638
As the calculated value of z lies within the critical region we reject the null hypothesis.
This relative frequency is an experimental probability: that of getting a red on a spinner. In 300 spins, we'd expect to get 0.4(300), or 120, reds.