let's firstly, convert the mixed fraction to improper fraction, and then subtract.
![\bf \stackrel{mixed}{2\frac{3}{4}}\implies \cfrac{2\cdot 4+3}{4}\implies \stackrel{improper}{\cfrac{11}{4}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{11}{4}-\cfrac{2}{3}\implies \stackrel{\textit{our LCD will be 12}}{\cfrac{(3)11-(4)2}{12}}\implies \cfrac{33-8}{12}\implies \cfrac{25}{12}\implies 2\frac{1}{12}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bmixed%7D%7B2%5Cfrac%7B3%7D%7B4%7D%7D%5Cimplies%20%5Ccfrac%7B2%5Ccdot%204%2B3%7D%7B4%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B11%7D%7B4%7D%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A%5Ccfrac%7B11%7D%7B4%7D-%5Ccfrac%7B2%7D%7B3%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bour%20LCD%20will%20be%2012%7D%7D%7B%5Ccfrac%7B%283%2911-%284%292%7D%7B12%7D%7D%5Cimplies%20%5Ccfrac%7B33-8%7D%7B12%7D%5Cimplies%20%5Ccfrac%7B25%7D%7B12%7D%5Cimplies%202%5Cfrac%7B1%7D%7B12%7D)
Correct Answer:Option A. 0.01
Solution:This is a problem of statistics and uses the concept of normal distributions. We need to convert the score of 90 into z-score and then find the desired probability from standard normal distribution table.
Converting 90 to z-score:

Now we are to find the probability of z score being more than 2.33. From the z-table the probability comes out to be 0.01.
Therefore, we can conclude that the probability of class average is greater than 90 is 0.01.
Answer:
1/9 = 0.11111111 ( infinite )
Step-by-step explanation:
24 total marbles
6 are black
so you have a 6/24, reduced to 1/4 probability of picking a black one.