Sum of two interiors=exterior




Answer: 0.22
Step-by-step explanation:
We know that the best point estimate for the difference between two population mean is the difference between their sample means.
Given : For the 39 randomly selected upperclassmen, the sample mean was 0.12 and sample standard deviation was 0.42. For the 35 randomly selected underclassmen, the sample mean was 0.34 and the sample standard deviation was 0.87.
Let A denotes the population of upperclassmen and B denotes the population of underclassmen .


Then, the point estimate of the difference in the population mean volunteered between underclassmen and upperclassmen will be :-

Hence, the point estimate of the difference in the population mean volunteered between underclassmen and upperclassmen =0.22
Answer:
The solutions are linearly independent because the Wronskian is not equal to 0 for all x.
The value of the Wronskian is 
Step-by-step explanation:
We can calculate the Wronskian using the fundamental solutions that we are provided and their corresponding the derivatives, since the Wroskian is defined as the following determinant.

Thus replacing the functions of the exercise we get:

Working with the determinant we get

Thus we have found that the Wronskian is not 0, so the solutions are linearly independent.
Answer:
A. 4(x - 4) + 2(3x² + 3x - 20)
C. (11x² + 7x - 55)-(5x² - 3x + 1)
F. (3x² + 5x - 28) + (3x² + 5x - 28)
Step-by-step explanation:
Given:
(3x-7)(2x+8)
= 6x² + 24x - 14x - 56
=6x² + 10x - 56
A. 4(x - 4) + 2(3x² + 3x - 20)
= 4x - 16 + 6x² + 6x - 40
= 6x² + 10x - 56
B. (3x² + 5x - 28) - (2x² + 4x + 28)
= 3x² + 5x - 28 - 2x² - 4x - 28
= x² + x - 56
C. (11x² + 7x - 55)-(5x² - 3x + 1)
= 11x² + 7x - 55 - 5x² + 3x - 1
= 6x² + 10x - 56
D. 4(x - 4) - 2(3x² + 3x - 20)
= 4x - 16 - 6x² - 6x + 40
= - 6x² - 2x + 24
E. (11x² + 7x - 55)-(5x² - 3x + 2)
= 11x² + 7x - 55 - 5x² + 3x - 2
= 11x² - 5x² + 7x + 3x - 55 - 2
= 6x² + 10x - 57
F. (3x² + 5x - 28) + (3x² + 5x - 28)
= 3x² + 5x - 28 + 3x² + 5x - 28
= 6x² + 10x - 56