Hello from MrBillDoesMath!
Answer: N = 143
Discussion:
This one took some trial and error! At first I listed all 2 digit primes, looked at the list, but didn't know how to proceed. So, I took the smallest 2 digit primes numbers: 11 and 13 and wondered if their product, 13*11 = 143, could be represented as the sum of 3 consecutive primes. I went back to my list of primes, added groups of three consecutive numbers that seemed to be in the right range to give the desired sum, and stumbled on 43, 47, and 53!
43 + 47 + 53 = 143 !
Therefore N = 143. It's the sum of 43, 47, and 53 as well as the product of 11 and 13.
Thank you,
MrB
It’s A. I hope that helps
Answer:
the zero property
Step-by-step explanation:
The answer to your question is b.
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.