Answer:
a) r = 0.974
b) Critical value = 0.602
Step-by-step explanation:
Given - Two separate tests are designed to measure a student's ability to solve problems. Several students are randomly selected to take both test and the results are give below
Test A | 64 48 51 59 60 43 41 42 35 50 45
Test B | 91 68 80 92 91 67 65 67 56 78 71
To find - (a) What is the value of the linear coefficient r ?
(b) Assuming a 0.05 level of significance, what is the critical value ?
Proof -
A)
r = 0.974
B)
Critical Values for the Correlation Coefficient
n alpha = .05 alpha = .01
4 0.95 0.99
5 0.878 0.959
6 0.811 0.917
7 0.754 0.875
8 0.707 0.834
9 0.666 0.798
10 0.632 0.765
11 0.602 0.735
12 0.576 0.708
13 0.553 0.684
14 0.532 0.661
So,
Critical r = 0.602 for n = 11 and alpha = 0.05
Answer:
Step-by-step explanation:
You should know that quotient is the result of division.
n/2
Four less than that would be subtracting.
(n/2) - 4 = 9
Answer:
(x - 9)(x + 3)
Step-by-step explanation:
Given
x² - 6x - 27
Consider the factors of the constant term (- 27) which sum to give the coefficient of the x- term (- 6)
The factors are - 9 and + 3, since
- 9 × 3 = - 27 and - 9 + 3 = - 6, thus
x² - 6x - 27 = (x - 9)(x + 3)
Answer:
I think it's 2566.8 if 4.6 is taken out of the account annually. I think that is what it is asking.. I'm only 13 so I'm not great at these yet but it should be right because I'm really good at math
Step-by-step explanation:
The coefficient would be c I believe I may be wrong