There are an infinite number of them.
-- Think of any number.
-- Multiply 4 by your number. Write it on top.
-- Multiply 5 by your number. Write it on the bottom.
-- You have a new fraction that's equivalent to 4/5 .
How many different numbers can you think of ?
That's how many new fractions you can create
that are ALL equivalent to 4/5 .
let's firstly convert the mixed fractions to improper fractions.
![\bf \stackrel{mixed}{3\frac{1}{2}}\implies \cfrac{3\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{7}{2}}~\hfill \stackrel{mixed}{1\frac{1}{3}}\implies \cfrac{1\cdot 3+1}{3}\implies \stackrel{improper}{\cfrac{4}{3}} \\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bmixed%7D%7B3%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B3%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B7%7D%7B2%7D%7D~%5Chfill%20%5Cstackrel%7Bmixed%7D%7B1%5Cfrac%7B1%7D%7B3%7D%7D%5Cimplies%20%5Ccfrac%7B1%5Ccdot%203%2B1%7D%7B3%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B4%7D%7B3%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D)
![\bf \begin{array}{ccll} miles&hours\\ \cline{1-2}\\ \frac{7}{2}&1\\[0.8em] m&\frac{4}{3} \end{array}\implies \cfrac{~~\frac{7}{2}~~}{m}=\cfrac{~~1~~}{\frac{4}{3}}\implies \cfrac{7}{2m}=\cfrac{3}{4}\implies 28=6m \\\\\\ \cfrac{28}{6}=m\implies \cfrac{14}{3}=m\implies 4\frac{2}{3}=m](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bccll%7D%20miles%26hours%5C%5C%20%5Ccline%7B1-2%7D%5C%5C%20%5Cfrac%7B7%7D%7B2%7D%261%5C%5C%5B0.8em%5D%20m%26%5Cfrac%7B4%7D%7B3%7D%20%5Cend%7Barray%7D%5Cimplies%20%5Ccfrac%7B~~%5Cfrac%7B7%7D%7B2%7D~~%7D%7Bm%7D%3D%5Ccfrac%7B~~1~~%7D%7B%5Cfrac%7B4%7D%7B3%7D%7D%5Cimplies%20%5Ccfrac%7B7%7D%7B2m%7D%3D%5Ccfrac%7B3%7D%7B4%7D%5Cimplies%2028%3D6m%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B28%7D%7B6%7D%3Dm%5Cimplies%20%5Ccfrac%7B14%7D%7B3%7D%3Dm%5Cimplies%204%5Cfrac%7B2%7D%7B3%7D%3Dm)
It’s just the same as if there were no decimal point. but if there isn’t a number in the quotient on the left of the decimal like #15, then you just place a 0 and continue to the next dividend which is 50.
We are choosing 2
2
r
shoes. How many ways are there to avoid a pair? The pairs represented in our sample can be chosen in (2)
(
n
2
r
)
ways. From each chosen pair, we can choose the left shoe or the right shoe. There are 22
2
2
r
ways to do this. So of the (22)
(
2
n
2
r
)
equally likely ways to choose 2
2
r
shoes, (2)22
(
n
2
r
)
2
2
r
are "favourable."
Another way: A perhaps more natural way to attack the problem is to imagine choosing the shoes one at a time. The probability that the second shoe chosen does not match the first is 2−22−1
2
n
−
2
2
n
−
1
. Given that this has happened, the probability the next shoe does not match either of the first two is 2−42−2
2
n
−
4
2
n
−
2
. Given that there is no match so far, the probability the next shoe does not match any of the first three is 2−62−3
2
n
−
6
2
n
−
3
. Continue. We get a product, which looks a little nicer if we start it with the term 22
2
n
2
n
. So an answer is
22⋅2−22−1⋅2−42−2⋅2−62−3⋯2−4+22−2+1.
2
n
2
n
⋅
2
n
−
2
2
n
−
1
⋅
2
n
−
4
2
n
−
2
⋅
2
n
−
6
2
n
−
3
⋯
2
n
−
4
r
+
2
2
n
−
2
r
+
1
.
This can be expressed more compactly in various ways.
The top and sides of the cylinder have an area of pr^2+2prh, but the pr^2 of the top is also removed from the cube which has an area of 6s^2.
A=6s^2+2prh
A=6*36+2p4*1
A=216+8p in^2
A≈241.13 in^2 (to nearest one-hundredth)
Note that I did not include the area of the cylinder that touches the cube...
