<span>0.41
There are several different ways to solve this problem. Since there's only 10 coins, you could simply calculate all 2^10 = 1024 possibilities using a spreadsheet and come up with the exact value of 416/1024 = 13/32 = 0.40625, or you can do it manually via reasoning. So:
Let's start with the quarters, and look at the situation of 0, 1, or 2 landing heads up.
0 - The value of all the dimes, nickels, and pennies add up to 39 cents, so Michael can't win.
2 - Got the 50 cents using quarters alone, so this is a 1 in 4 chance. So we have 0.25
1 - This is where it's interesting. And there's a 2 in 4 or 1/2 chance of 1 quarter coming up heads. In this case, we need at least 25 cents worth for the other 8 coins. So let's break down this case.
Let's look at the issue of 0, 1, 2, or 3 dimes.
0 - The nickel and pennies add up to 9, so we don't win.
1 - The nickel and pennies add up to 9, so our total is 19, so we don't win.
3 - We have 30 cents and with the 25 from the quarter we're good to go. This happens 1 out of 8 times. So 1/8 * 1/2 = 1/16 = 0.0625. Adding to the 0.25 gives us 0.3125
2 - We have 20 cents, plus the 25 from the quarter. So we need 5 more cents. We get to this situation 3/8 * 1/2 = 3/16 times.
Nickle
0 = Pennies only add to 4, so can't win.
1 = Gives us the 50 cents we need. So 1/2*3/16 = 3/32 = 0.09375. Adding that to the 0.3125 we already have gives us 0.40625 which exactly matches the exhaustive search of all 1024 possibilities.</span>
Answer:
4
Step-by-step explanation:
Because there are 10 pices and 4 of them are greater than 6
It would be (x−9)(x+4)<span>
</span>
as you already know, to get the inverse of any expression we start off by doing a quick switcheroo on the variables and then solving for "y", let's do so.
![\stackrel{f(x)}{y}~~ = ~~\cfrac{2x}{x+3}\implies \stackrel{\textit{quick switcheroo}}{x~~ = ~~\cfrac{2y}{y+3}}\implies xy+3x=2y\implies xy-2y+3x=0 \\\\\\ xy-2y=-3x\implies \stackrel{\textit{common factoring}}{y(x-2)=-3x}\implies y=\cfrac{-3x}{x-2}\implies \stackrel{f^{-1}(x)}{y}=\cfrac{3x}{2-x}](https://tex.z-dn.net/?f=%5Cstackrel%7Bf%28x%29%7D%7By%7D~~%20%3D%20~~%5Ccfrac%7B2x%7D%7Bx%2B3%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bquick%20switcheroo%7D%7D%7Bx~~%20%3D%20~~%5Ccfrac%7B2y%7D%7By%2B3%7D%7D%5Cimplies%20xy%2B3x%3D2y%5Cimplies%20xy-2y%2B3x%3D0%20%5C%5C%5C%5C%5C%5C%20xy-2y%3D-3x%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bcommon%20factoring%7D%7D%7By%28x-2%29%3D-3x%7D%5Cimplies%20y%3D%5Ccfrac%7B-3x%7D%7Bx-2%7D%5Cimplies%20%5Cstackrel%7Bf%5E%7B-1%7D%28x%29%7D%7By%7D%3D%5Ccfrac%7B3x%7D%7B2-x%7D)
Answer:
![6\sqrt{2}+2{\sqrt3}](https://tex.z-dn.net/?f=6%5Csqrt%7B2%7D%2B2%7B%5Csqrt3%7D)
Step-by-step explanation:
Given expression:
![\sqrt{72}-3\sqrt{12}+\sqrt{192}](https://tex.z-dn.net/?f=%5Csqrt%7B72%7D-3%5Csqrt%7B12%7D%2B%5Csqrt%7B192%7D)
Rewrite 72 as (36 · 2), 12 as (4 · 3), and 192 as (64 · 3):
![\implies \sqrt{36 \cdot 2}-3\sqrt{4 \cdot 3}+\sqrt{64 \cdot 3}](https://tex.z-dn.net/?f=%5Cimplies%20%5Csqrt%7B36%20%5Ccdot%202%7D-3%5Csqrt%7B4%20%5Ccdot%203%7D%2B%5Csqrt%7B64%20%5Ccdot%203%7D)
Apply the radical rule
:
![\implies \sqrt{36}\sqrt{2}-3\sqrt{4}\sqrt{3}+\sqrt{64}{\sqrt3}](https://tex.z-dn.net/?f=%5Cimplies%20%5Csqrt%7B36%7D%5Csqrt%7B2%7D-3%5Csqrt%7B4%7D%5Csqrt%7B3%7D%2B%5Csqrt%7B64%7D%7B%5Csqrt3%7D)
Rewrite 36 as 6², 4 as 2², and 64 as 8²:
![\implies \sqrt{6^2}\sqrt{2}-3\sqrt{2^2}\sqrt{3}+\sqrt{8^2}{\sqrt3}](https://tex.z-dn.net/?f=%5Cimplies%20%5Csqrt%7B6%5E2%7D%5Csqrt%7B2%7D-3%5Csqrt%7B2%5E2%7D%5Csqrt%7B3%7D%2B%5Csqrt%7B8%5E2%7D%7B%5Csqrt3%7D)
Apply the radical rule
:
![\implies 6\sqrt{2}-3\cdot 2\sqrt{3}+8{\sqrt3}](https://tex.z-dn.net/?f=%5Cimplies%206%5Csqrt%7B2%7D-3%5Ccdot%202%5Csqrt%7B3%7D%2B8%7B%5Csqrt3%7D)
Simplify:
![\implies 6\sqrt{2}-6\sqrt{3}+8{\sqrt3}](https://tex.z-dn.net/?f=%5Cimplies%206%5Csqrt%7B2%7D-6%5Csqrt%7B3%7D%2B8%7B%5Csqrt3%7D)
![\implies 6\sqrt{2}+2{\sqrt3}](https://tex.z-dn.net/?f=%5Cimplies%206%5Csqrt%7B2%7D%2B2%7B%5Csqrt3%7D)