Answer:
B. The variable s represents the number of students in each class.
C. The coefficient 14 represents the number of classes in the 9th grade.
Step-by-step explanation:
The total number can be found by multiplying the number of the things with the number of people producing it. For example if 5 boys make 5 colored ropes the total number of ropes will be 5*5= 25.
In this question the combinations rule is used for variuos number of classes which are 14. Now we have to find the number of students which are s (s-1). Suppose s= 6 so the number of students would be 6(6-1) = 6(5) = 30
We will multiply the number of classes 14 with the number of students s(s-1) to get the total number of cards produced.
X*y=72
x = 5y-2
x=72/y
72/y=5y-2
72=5y^2-2y
5y^2 -2y - 72 = 0
(5y+18)(y-4)=0
y = -18/5 discard
or y = 4
4x=72
x=18
(18,4)
X=32
Explanation:
When you change sides, if in the left side you are multiplying, when you transfer it to right side you divide. Same thing should be done for addition subtraction.
So /-4 was division and negative number
When we transfered it, it became positive multiplication number
Answer:
0.2514 = 25.14% probability that the diameter of a selected bearing is greater than 85 millimeters.
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 question, we have that:

Find the probability that the diameter of a selected bearing is greater than 85 millimeters.
This is 1 subtracted by the pvalue of Z when X = 85. Then



has a pvalue of 0.7486.
1 - 0.7486 = 0.2514
0.2514 = 25.14% probability that the diameter of a selected bearing is greater than 85 millimeters.