For each student in a certain class, a teacher adjusted the student's test score using the formula y = 0.8x + 20, where x is the student's original test score and y is the student's adjusted test score. if the standard deviation of the original test scores of the students in the class was 20, what was the standard deviation of the adjusted test scores of the students in the class? a. 12 b. 16 c. 28 d. 36 e. 40

Option B) 16

Step-by-step explanation:

We are given the following in the question:

where x is the student's original test score and y is the student's adjusted test score.

We have to find the standard deviation of the adjusted test scores of the students in the class, if the standard deviation of the original test scores of the students in the class was 20.

We know that:

Thus, if we add 20 to each data set then, the standard deviation does not change.

But multiplying each score by 0.8, changes the standard deviation 0.8 times.

Thus, we can write:

Thus, standard deviation of adjusted score is 16.

i can you, i love poetry

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

cretdit to source