Answer:
14.63% probability that a student scores between 82 and 90
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:

What is the probability that a student scores between 82 and 90?
This is the pvalue of Z when X = 90 subtracted by the pvalue of Z when X = 82. So
X = 90



has a pvalue of 0.9649
X = 82



has a pvalue of 0.8186
0.9649 - 0.8186 = 0.1463
14.63% probability that a student scores between 82 and 90
Answer:
rectangle
kite
Step-by-step explanation:
Answer:x=3
Step-by-step explanation:
D( x )
x+2 = 0
x = 0
x+2 = 0
x+2 = 0
x+2 = 0 // - 2
x = -2
x = 0
x = 0
x in (-oo:-2) U (-2:0) U (0:+oo)
(9*x-7)/(x+2)+15/x = 9 // - 9
(9*x-7)/(x+2)+15/x-9 = 0
(x*(9*x-7))/(x*(x+2))+(15*(x+2))/(x*(x+2))+(-9*x*(x+2))/(x*(x+2)) = 0
x*(9*x-7)+15*(x+2)-9*x*(x+2) = 0
9*x^2-9*x^2+8*x-18*x+30 = 0
30-10*x = 0
(30-10*x)/(x*(x+2)) = 0
(30-10*x)/(x*(x+2)) = 0 // * x*(x+2)
30-10*x = 0
30-10*x = 0 // - 30
-10*x = -30 // : -10
x = -30/(-10)
x = 3
x = 3
Take 47,000/12
12 months in a year = $3917
$3917.00 is what he would take home every month before taxes of course.
Answer:
"B A C" 16x20=320
Step-by-step explanation: