Answer:
true
Step-by-step explanation:

Answer:
(a) Margin of error ( E) = $2,000 , n = 54
(b) Margin of error ( E) = $1,000 , n = 216
(c) Margin of error ( E) = $500 , n= 864
Step-by-step explanation:
Given -
Standard deviation
= $7,500
= 1 - confidence interval = 1 - .95 = .05
=
= 1.96
let sample size is n
(a) Margin of error ( E) = $2,000
Margin of error ( E) = 
E = 
Squaring both side


n = 54.0225
n = 54 ( approximately)
(b) Margin of error ( E) = $1,000
E = 
1000 = 
Squaring both side


n = 216
(c) Margin of error ( E) = $500
E = 
500 = 
Squaring both side


n = 864
Larry needs
to score
33 points in the last game to have an average of 25 PPG.
To find the answer, we can solve the equation:

<em />(where <em>p</em> is how many points he needs to score in the last game)
To find the mean of a group of numbers, you add them all up and divide the total by the number of numbers there are. Since Larry averaged 23 PPG in 4 games, we can multiply 23 by 4 to get the total of the first 4 games from the data. Then, we find <em>p</em>, which we will add to get our final total. Then, you divide by the 5 games.




First, I simplified 4 x 23 to get 92. Then, I multiplied each side by 5 to get rid of the denominator. Finally, I subtracted 92 from each side to isolate <em>p</em>, and found that <em>p</em> = 33.
That is approximately, in order, a C, a C, a B, a D, an A, a B, and an A.
In grade points that is a 2, a 2, a 3, a 1, a 4, a 3, and a 4.
The average of those numbers is about 2.7, so you have a 2.7.
You can raise that by bringing up the D as it is an outlier here.
Answer:
3/4
Step-by-step explanation:
divided by 3 to get the answer.