8n−7y−12n+5−3y
=8n+−7y+−12n+5+−3y
Combine Like Terms:
=8n+−7y+−12n+5+−3y
=(8n+−12n)+(−7y+−3y)+(5)
=−4n+−10y+5
Correct Ans:Option A. 0.0100
Solution:We are to find the probability that the class average for 10 selected classes is greater than 90. This involves the utilization of standard normal distribution.
First step will be to convert the given score into z score for given mean, standard deviation and sample size and then use that z score to find the said probability. So converting the value to z score:

So, 90 converted to z score for given data is 2.326. Now using the z-table we are to find the probability of z score to be greater than 2.326. The probability comes out to be 0.01.
Therefore, there is a 0.01 probability of the class average to be greater than 90 for the 10 classes.
In order to solve or know the probability of having 2 girls
and 2 boys, assumed that a girl is as likely as a girl at each birth, pascal’s
triangle will be likely used. And we will be referring to the line 4 of pascal’s
triangle, which was 1 4 6 4 1. Then it
will look like this: 1 = 4 girls; 4 = 3 girls & 1 boy; 6 = 2 girls & 2
boys; 4 = 3 boys & 1 girl; 1 = 4 boys. And now for the solution in order to
get the probability of having 2 girls and 2 boys is to divided into the sum of 1+4+6+4+1.
Step-by-step explanation:
10300000 = 1,03 * 10⁷