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3 years ago

It’s My Final Test I’m not sure what the answer is

Mathematics

2 answers:

cestrela7 [59]3 years ago

7 0

Answer:

The answer is B .

Step-by-step explanation

Firstly

= (8 - √2 ) × ( 9 + √5 ) (writing the given only )

= 8 ( 9 + √5) - √2 ( 9+√5 ) ( multiplying them )

= 72 + 8√5 - 9√2 - √10

Hope you got it now

Elena-2011 [213]3 years ago

4 0

Answer:

Step-by-step explanation:

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3 years ago

For about $1 billion in new space shuttle expenditures, NASA has proposed to install new heat pumps, power heads, heat exchanger

11111nata11111 [884]

Answer:

The probability of one or more catastrophes in:

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(2) Five mission is 0.0410.

(3) Ten mission is 0.0803.

(4) Fifty mission is 0.3419.

Step-by-step explanation:

Let X = number of catastrophes in the missions.

The probability of a catastrophe in a mission is, P (X) = $p=\frac{1}{120}$ .

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X is:

$P(X=x)={n\choose x}\frac{1}{120}^{x}(1-\frac{1}{120})^{n-x};\x=0,1,2,3...$

In this case we need to compute the probability of 1 or more than 1 catastrophes in n missions.

Then the value of P (X ≥ 1) is:

P (X ≥ 1) = 1 - P (X < 1)

= 1 - P (X = 0)

$=1-{n\choose 0}\frac{1}{120}^{0}(1-\frac{1}{120})^{n-0}\\=1-(1\times1\times(1-\frac{1}{120})^{n-0})\\=1-(1-\frac{1}{120})^{n-0}$

(1)

Compute the compute the probability of 1 or more than 1 catastrophes in 2 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{2-0}=1-0.9834=0.0166$

(2)

Compute the compute the probability of 1 or more than 1 catastrophes in 5 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{5-0}=1-0.9590=0.0410$

(3)

Compute the compute the probability of 1 or more than 1 catastrophes in 10 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{10-0}=1-0.9197=0.0803$

(4)

Compute the compute the probability of 1 or more than 1 catastrophes in 50 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{50-0}=1-0.6581=0.3419$

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3 years ago

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Answer:

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3 years ago

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