Answer:
30 is the lest common multiple
Step-by-step explanation:
Answer:
0.1507 or 15.07%.
Step-by-step explanation:
We have been given that the manufacturing of a ball bearing is normally distributed with a mean diameter of 22 millimeters and a standard deviation of .016 millimeters. To be acceptable the diameter needs to be between 21.97 and 22.03 millimeters.
First of all, we will find z-scores for data points using z-score formula.
, where,
z = z-score,
x = Sample score,
= Mean,
= Standard deviation.



Let us find z-score of data point 22.03.



Using probability formula
, we will get:

Therefore, the probability that a randomly selected ball bearing will be acceptable is 0.1507 or 15.07%.
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)
Answer:
6 gumballs can be stored in the gumball machine.
Step-by-step explanation:
Since a gumball machine consists of a base and a glass sphere, and the gumballs are stored in the glass sphere, if each gumball has a diameter of 1.5 inches, to determine how many gumballs can be stored in a gumball machine whose glass sphere measures 10 inches in diameter, the following calculation must be made:
10 / 1.5 = X
6.666666 = X
Thus, 6 gumballs can be stored in the gumball machine.