Answer:
1/15 each
Step-by-step explanation:
1/5 of chocolate is separated into 3
1/5 divided by 3 = 1/5 x 1/3 = 1/15 of the whole bar
Check the picture below.
let's firstly convert the mixed fractions to improper fractions.
![\stackrel{mixed}{7\frac{1}{2}}\implies \cfrac{7\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{15}{2}} ~\hfill \stackrel{mixed}{12\frac{1}{2}}\implies \cfrac{12\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{25}{2}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cstackrel%7Bmixed%7D%7B7%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B7%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B15%7D%7B2%7D%7D%20~%5Chfill%20%5Cstackrel%7Bmixed%7D%7B12%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B12%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B25%7D%7B2%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\stackrel{\textit{\Large Areas}}{\stackrel{two~triangles}{2\left[ \cfrac{1}{2}\left(\cfrac{15}{2} \right)(10) \right]}~~ + ~~\stackrel{\textit{three rectangles}}{(10)(15)~~ + ~~\left( \cfrac{15}{2} \right)(15)~~ + ~~\left( \cfrac{25}{2} \right)(15)}} \\\\\\ 75~~ + ~~150~~ + ~~112.5~~ + ~~187.5\implies \boxed{525}](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Ctextit%7B%5CLarge%20Areas%7D%7D%7B%5Cstackrel%7Btwo~triangles%7D%7B2%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%5Cleft%28%5Ccfrac%7B15%7D%7B2%7D%20%5Cright%29%2810%29%20%5Cright%5D%7D~~%20%2B%20~~%5Cstackrel%7B%5Ctextit%7Bthree%20rectangles%7D%7D%7B%2810%29%2815%29~~%20%2B%20~~%5Cleft%28%20%5Ccfrac%7B15%7D%7B2%7D%20%5Cright%29%2815%29~~%20%2B%20~~%5Cleft%28%20%5Ccfrac%7B25%7D%7B2%7D%20%5Cright%29%2815%29%7D%7D%20%5C%5C%5C%5C%5C%5C%2075~~%20%2B%20~~150~~%20%2B%20~~112.5~~%20%2B%20~~187.5%5Cimplies%20%5Cboxed%7B525%7D)
Answer:
A.
Step-by-step explanation:
Answer:
The answer is 350 points
Step-by-step explanation:
The Area of a circle is πr∧2
The problem states that in the attempts, you never hit the outermost ring in the 10 attempts so we need the area of the 1cm and 2cm circles
Area of the 1st circle; π X 1∧2 = π
Area of the Second circle; π X 2∧2 = 4π
We also need the area which is the difference between the area of the 1cm and 2cm circle
4π - π = π
Points 50 = π/4π = 1/4
points 30 = 3π/4π = 3/4
For one attempt,
E(x) = 50 x 1/4 + 30 x 3/4
= 12.25 + 22.5
= 34.75
This is approximately 35
Therefor, 10 individual attempts will be 10 x 35 = 350
Answer:
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Step-by-step explanation:
