Proving the Parallelogram Side Theorem

Mathematics

1 answer:

zzz [600]3 years ago

4 0

Answer:

Statements Reasons

1. ABCD is a parallelogram 1. Given

2. Seg. AB is parallel to seg. CD 2. Def of parallelogram

3. Seg. AD is parallel to seg. BC 3. Def of parallelogram

4. Draw seg. BD 4. Between 2 pts there is 1 line

5. <ABD is congr <CDB 5. Alt. Int. < Theorem

6. <BDA is cong <DBC 6. Alt. Int. < Theorem

7. Seg. BD is congr seg. BD 7. Congr. of seg's is reflexive

8. Triangle BAD is congr tr. DCB 8. SAS Postulate

9. AB is congr CD 9. CPCTC

10. BC is congr DA 10. CPCTC

11. AB = CD; BC = DA 11. Def of congr. seg's

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The probability that a male will be colorblind is .042. Find the probabilities that in a group of 53 men, the following are true

Dvinal [7]

Answer:

See below for answers and explanations

Step-by-step explanation:

Part A

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

Part B

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