let's firstly, convert the mixed fractions to improper, and then do equation.
![\bf \stackrel{mixed}{3\frac{4}{5}}\implies \cfrac{3\cdot 5+4}{5}\implies \stackrel{improper}{\cfrac{19}{5}} ~\hfill \stackrel{mixed}{2\frac{5}{7}}\implies \cfrac{2\cdot 7+5}{7}\implies \stackrel{improper}{\cfrac{19}{7}} \\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%5Cbf%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%0A~%5Chfill%0A%5Cstackrel%7Bmixed%7D%7B2%5Cfrac%7B5%7D%7B7%7D%7D%5Cimplies%20%5Ccfrac%7B2%5Ccdot%207%2B5%7D%7B7%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B19%7D%7B7%7D%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D)
Answer:
Permutation is used.
No. of arrangements = 1260
Step-by-step explanation:
Permutation is used for choosing & arrangement of 'r' out of 'n' objects, when sequence is important.
Total no. of alphabets in CORRECT = 7
C is repeated twice, R is repeated twice
So, Total No. of arrangements = <u>7 !</u>
(2!)(2!)
= 5040 / 4
= 1260
Answer:
-9f
Step-by-step explanation:
Answer:
Katie is not cheating. From the question, it sounds like Katie's spinner is 1-12. James has two dice so he has two sets of each number 1-6. This means that Katie can roll two numbers greater than six but James can only roll up to six. Using probability, I think there is a 50% chance that Katie's number will be bigger than James every time she spins. A fairer way would be for them both to play with the same dice or spinner.
Step-by-step explanation:
