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2 years ago

I can't figure this out

Mathematics

2 answers:

Lerok [7]2 years ago

6 0

10x22 and then divide by 3.14
So the answer is 70

EastWind [94]2 years ago

3 0

I need help with this same problem too , can someone please help us be

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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:

(1) Two mission is 0.0166.

(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:

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

6 0

3 years ago

16 ounces is equivalent to 1 pound. A lion cub born at

Musya8 [376]

Answer:

368 ounces

Step-by-step explanation:

since 16 ounces is 1 pound, multiply 23 by 16

5 0

3 years ago

Read 2 more answers

Find the constant of variation for the relationship shown in the following table:

EastWind [94]

In order to get the constant of variation, you can either make a linear equation that relates x and y in which the slope is the constant of variation

LINEARIZING

x = 3y

the slope of the equation is 3 and therefore the constant of variation is 3

5 0

3 years ago

Solve –vp + 40 < 95 for v. Show your work.

vfiekz [6]

The answer is v>-55/p
When dividing or multiplying an inequality by a negative, the sign changes. If it was > it becomes <. If it was >= it becomes <=. Hope I was able to help

4 0

3 years ago

The assumption of exponentially distributed service times indicates that a. service time increase at an exponential rate as the

OleMash [197]

Answer:

option c is correct.

Step-by-step explanation:

Exponential distribution is large class of probability that includes its member .It also include normal distribution, binomial , gamma , Poisson and many other distribution as well.

Example

It is concerned with the amount of time until any event occur.For example length , in minutes and long distance telephone,

so from given statement it is about time service so option c 63% of the service are less than the mean service is true about it

4 0

3 years ago