I Belive That There Can't Be 3 Numbers In A Rectangle Because At The Top You Get 24 And 6 And I Think It Is BxH One Half Times Base Times Height So 2xBxH .
Answer:
a
Step-by-step explanation:
Answer:
34 Braids
Step-by-step explanation:
.85 per braid means 1 is 85 cents, 2 is $1.70 and so on. So normally you would just divide the total amount of money, so for instance if you had $2.55 you would be able to afford 3 braids, since 3 braids cost $2.55. in fact $2.56, $2.57 aaaaaaaaaaall the way up to $3.39 would be 3 braids, then $3.40 would be 4 braids.
Here though you first pay 3 dollars. so you need 3 first, then an extra 85 cents to get 1 braid. 2 braids is 3 dollars plus 1.70, 3 braids is 3 + 2.55 4 braids is 3 + 3.40 and so on.
So what you want to do is first take away that 3 dollars you need before the braids, then divide by .85. for instance $6.40 would get you 4 braids as I showed bcause 3 + 3.40 = 6.40. now 6.40 - 3 = 3.40 then 3.40 divided by .85 = 4
The problem says we start with$32, so lets do those steps. first 32 - 3 = 29. then 29 divided by .85 = 34.11. well it's not whole number so lets try something. 34 braids would be 3 + .85*34 = 31.90 and 35 braids would be 32.75 So 31.90, 31.91 aaaaaaaaall the way up to 32.74 would be 34 braids then 32.75 would be 35 braids. well, $32 is in that in between space, so it is 34 braids.
When I did 32 - 3 = 29. then 29 divided by .85 = 34.11 that 34.11 does actually tell us we wouldn't get to 35, I just wanted to make it very clear. so the answer is 34 braids.
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.
