let's firstly convert the mixed fractions to improper fractions and then divide.
![\bf \stackrel{mixed}{1\frac{1}{4}}\implies \cfrac{1\cdot 4+1}{4}\implies \stackrel{improper}{\cfrac{5}{4}}~\hfill \stackrel{mixed}{3\frac{4}{5}}\implies \cfrac{3\cdot 5+4}{5}\implies \stackrel{improper}{\cfrac{19}{5}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{5}{4}\div\cfrac{19}{5}\implies \cfrac{5}{4}\cdot \cfrac{5}{19}\implies \cfrac{25}{76}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bmixed%7D%7B1%5Cfrac%7B1%7D%7B4%7D%7D%5Cimplies%20%5Ccfrac%7B1%5Ccdot%204%2B1%7D%7B4%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B5%7D%7B4%7D%7D~%5Chfill%20%5Cstackrel%7Bmixed%7D%7B3%5Cfrac%7B4%7D%7B5%7D%7D%5Cimplies%20%5Ccfrac%7B3%5Ccdot%205%2B4%7D%7B5%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B19%7D%7B5%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B5%7D%7B4%7D%5Cdiv%5Ccfrac%7B19%7D%7B5%7D%5Cimplies%20%5Ccfrac%7B5%7D%7B4%7D%5Ccdot%20%5Ccfrac%7B5%7D%7B19%7D%5Cimplies%20%5Ccfrac%7B25%7D%7B76%7D)
Answer:
adding by 5
Step-by-step explanation:
,...............................
4x=20-8y
x=5-2y now use this value of x in the second equation...
3(5-2y)+6y=15
15-6y+6y=15
15=15 This is true for any y or x value.
So there are infinitely many solutions as the two equations describe the same line.
Answer:
The minimum sample size required to ensure that the estimate has an error of at most 0.14 at the 95% level of confidence is n=567.
Step-by-step explanation:
We have to calculate the minimum sample size n needed to have a margin of error below 0.14.
The critical value of z for a 95% confidence interval is z=1.96.
To do that, we use the margin of error formula in function of n:

The minimum sample size to have this margin of error is n = 567.
2 Letters chosen among 6 with no repeat:
a) the first has 6 ways
the second has 5 ways
Then to choose 2 letters among 6, with no repeat, there are 6x5 = 30 ways
b) 3 numbers are taken from the digits 0 through 9 with repeats (total = 10)
the first has 10 ways
the second has 10 ways
the third has 10 ways
Then to choose 3 numbers among 10, with repeat, there are 10x10x10= 1000 ways
And the total serial numbers that can be generated is
30 x 1000 = 3000