x = total amount split between Adam and Tom.
since we know the total amount split between both in a 18 : 17 ratio is "x", let's divide "x" by (18 + 17) and distribute accordingly to get the amount of each.
![\stackrel{Adam~received}{18\cdot \cfrac{x}{18+17}}\qquad \qquad \stackrel{Tom~received}{17\cdot \cfrac{x}{18+17}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{since we know that Adam got "5" more}}{ \stackrel{Adam}{18\cdot \cfrac{x}{18+17}}~~ = ~~\stackrel{Tom}{17\cdot \cfrac{x}{18+17}~~ + ~~5} }\qquad \implies \qquad \cfrac{18x}{35}~~ + ~~\cfrac{17x}{35}+5](https://tex.z-dn.net/?f=%5Cstackrel%7BAdam~received%7D%7B18%5Ccdot%20%5Ccfrac%7Bx%7D%7B18%2B17%7D%7D%5Cqquad%20%5Cqquad%20%5Cstackrel%7BTom~received%7D%7B17%5Ccdot%20%5Ccfrac%7Bx%7D%7B18%2B17%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bsince%20we%20know%20that%20Adam%20got%20%225%22%20more%7D%7D%7B%20%5Cstackrel%7BAdam%7D%7B18%5Ccdot%20%5Ccfrac%7Bx%7D%7B18%2B17%7D%7D~~%20%3D%20~~%5Cstackrel%7BTom%7D%7B17%5Ccdot%20%5Ccfrac%7Bx%7D%7B18%2B17%7D~~%20%2B%20~~5%7D%20%7D%5Cqquad%20%5Cimplies%20%5Cqquad%20%5Ccfrac%7B18x%7D%7B35%7D~~%20%2B%20~~%5Ccfrac%7B17x%7D%7B35%7D%2B5)
![\stackrel{\textit{multiplying both sides by }\stackrel{LCD}{35}}{35\left( \cfrac{18x}{35} \right)~~ = ~~35\left( \cfrac{17x}{35}+5 \right)}\implies 18x~~ = ~~17x+175\implies \boxed{x =175} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{Tom~received}{17\cdot \cfrac{x}{18+17}}\implies \cfrac{17(175)}{35}\implies \blacktriangleright 85 \blacktriangleleft](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%20%7D%5Cstackrel%7BLCD%7D%7B35%7D%7D%7B35%5Cleft%28%20%5Ccfrac%7B18x%7D%7B35%7D%20%5Cright%29~~%20%3D%20~~35%5Cleft%28%20%5Ccfrac%7B17x%7D%7B35%7D%2B5%20%5Cright%29%7D%5Cimplies%2018x~~%20%3D%20~~17x%2B175%5Cimplies%20%5Cboxed%7Bx%20%3D175%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7BTom~received%7D%7B17%5Ccdot%20%5Ccfrac%7Bx%7D%7B18%2B17%7D%7D%5Cimplies%20%5Ccfrac%7B17%28175%29%7D%7B35%7D%5Cimplies%20%5Cblacktriangleright%2085%20%5Cblacktriangleleft)
Answer:
Step-by-step explanation:
Let n be a random variable that represents the first Jonathan apple chosen at random that has bitter pit.
a) P(X = n) = q(n-1)p, where q = 1 - p.
From the information given, probability if success, p = 12.6/100 = 0.126
b) for n = 3, the probability value from the geometric probability distribution calculator is
P(n = 3) = 0.096
For n = 5, the probability value from the geometric probability distribution calculator is
P(n = 5) = 0.074
For n = 12, the probability value from the geometric probability distribution calculator is
P(n = 12) = 0.8
c) For n ≥ 5, the probability value from the geometric probability distribution calculator is
P(n ≥ 5) = 0.58
d) the expected number of apples that must be examined to find the first one with bitter pit is the mean.
Mean = 1/p
Mean = 1/0.126 = 7.9
Approximately 8 apples
We cannot solve for a numerical value of p, but rather set the equation equal to p.
w = 4/(p+2)
(Multiply both sides by the denominator)
w(p+2) = 4
(divide both sides by w)
p+2 = 4/w
(Subtract 2 from both sides) (also 2/1)
p = (4-2)/w
p = 2/w
