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Romashka-Z-Leto [24]
3 years ago
8

Can someone help me solve #1 and #2 by working backward? Here is a picture:

Mathematics
1 answer:
Brilliant_brown [7]3 years ago
3 0
1. 4(n+3)=16, n+3=4, n=1
2. m/4+3=24, m/4=21, m=84
1. three added to a number n implies n+3, 4 times that sum is 4 (n+3), to solve divide both sides by 4 then subtract three from both sides.
2. a number divided by 4 means n/4 adding it to three n/4+3, to solve subtract 3 from both sides then multiply both sides by 4
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<u>Part A</u>

\displaystyle P(X=x)=\binom{n}{x}p^xq^{n-x}\\\\P(X=5)=\binom{53}{5}(0.042)^5(1-0.042)^{53-5}\\\\P(X=5)=\frac{53!}{(53-5)!*5!}(0.042)^5(0.958)^{48}\\\\P(X=5)\approx0.0478

<u>Part B</u>

P(X\leq5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)\\\\P(X\leq5)=\binom{53}{1}(0.042)^1(1-0.042)^{53-1}+\binom{53}{2}(0.042)^2(1-0.042)^{53-2}+\binom{53}{3}(0.042)^3(1-0.042)^{53-3}+\binom{53}{4}(0.042)^4(1-0.042)^{53-4}+\binom{53}{1}(0.042)^5(1-0.042)^{53-5}\\\\P(X\leq5)\approx0.9767

<u>Part C</u>

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