We have the frequencies for each of the grades. We can estimate the number of students graded by adding all those frequencies. Let's call N the total number of grades:
We have then a total number of grades of 39.
The corresponding relative frequency for a grade is the ratio of the frequency to the total number of "samples", 39 in this case.
Then, for grade A, the relative frequency (RF) will be:
This will be the fraction of the total grades that are A. Represented as a percentage will be 10.26%, rounded to two decimal places.
Now, to complete the table we do the same for the other frequencies:
For grade B:
For grade C:
For grade D:
For grade F:
Answer:
Step-by-step explanation:
h(x+1) = ( x + 1 - 2)²
= (x -1)²
Simplify the following:
((x^2 - 11 x + 30) (x^2 + 6 x + 5))/((x^2 - 25) (x - 5 x - 6))
The factors of 5 that sum to 6 are 5 and 1. So, x^2 + 6 x + 5 = (x + 5) (x + 1):
((x + 5) (x + 1) (x^2 - 11 x + 30))/((x^2 - 25) (x - 5 x - 6))
The factors of 30 that sum to -11 are -5 and -6. So, x^2 - 11 x + 30 = (x - 5) (x - 6):
((x - 5) (x - 6) (x + 5) (x + 1))/((x^2 - 25) (x - 5 x - 6))
x - 5 x = -4 x:
((x - 5) (x - 6) (x + 5) (x + 1))/((x^2 - 25) (-4 x - 6))
Factor -2 out of -4 x - 6:
((x - 5) (x - 6) (x + 5) (x + 1))/(-2 (2 x + 3) (x^2 - 25))
x^2 - 25 = x^2 - 5^2:
((x - 5) (x - 6) (x + 5) (x + 1))/(-2 (x^2 - 5^2) (2 x + 3))
Factor the difference of two squares. x^2 - 5^2 = (x - 5) (x + 5):
((x - 5) (x - 6) (x + 5) (x + 1))/(-2(x - 5) (x + 5) (2 x + 3))
((x - 5) (x - 6) (x + 5) (x + 1))/((x - 5) (x + 5) (-2) (2 x + 3)) = ((x - 5) (x + 5))/((x - 5) (x + 5))×((x - 6) (x + 1))/(-2 (2 x + 3)) = ((x - 6) (x + 1))/(-2 (2 x + 3)):
((x - 6) (x + 1))/(-2 (2 x + 3))
Multiply numerator and denominator of ((x - 6) (x + 1))/(-2 (2 x + 3)) by -1:
Answer: (-(x - 6) (x + 1))/(2 (2 x + 3))
Answer:
X=-22
Step-by-step explanation:
