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3 years ago

Hi, can someone help me on these 2 as quick as possible please, ty!!

Mathematics

2 answers:

Advocard [28]3 years ago

8 0

1. a) true

1. b) false

1. c) false

1. d) true

2. Sasha could have done a reflection where x = 1 to have done only one transformation.

tensa zangetsu [6.8K]3 years ago

4 0

Answer:

1A T

1B F

1C F

1D F

2: Sasha reflected across x= -1

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2 years ago

For what value of p will the pair of linear equations have infinitely many solutions (p-3)x + 3y = p , px+ py = 12

sleet_krkn [62]

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Step-by-step explanation:

HI !

px + 3y - ( p-3)=0

a₁ = p , b₁ = 3 , c₁ = - (p-3)

=====================

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a₂ = 12 , b₂ = p , c₂ = -p

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--------------------------------------

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2 years ago

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

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

Part C

$\displaystyle P(X\geq 1)=1-P(X=0)\\\\P(X\geq1)=1-(1-0.042)^{53}\\\\P(X\geq1)\approx1-0.1029\\\\P(X\geq1)\approx0.8971$

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2 years ago