X 5. 2, for x = 10<br> Pls give answer!!!!

The probability that a single radar station will detect an enemy plane is 0.65.

taurus [48]

Answer:

a) We need 4 stations to be 98% certain that an enemy plane flying over will be detected by at least one station.

b) If seven stations are in use, the expected number of stations that will detect an enemy plane is 4.55.

Step-by-step explanation:

For each station, there are two two possible outcomes. Either they detected the enemy plane, or they do not. This means that we can solve this problem using concepts of the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem, we have that:

The probability that a single radar station will detect an enemy plane is 0.65. This means that $n = 0.65$ .

(a) How many such stations are required to be 98% certain that an enemy plane flying over will be detected by at least one station?

This is the value of n for which $P(X = 0) \leq 0.02$ .

n = 1.

$P(X = 0) = C_{1,0}.(0.65)^{0}.(0.35)^{1} = 0.35$

n = 2

$P(X = 0) = C_{2,0}.(0.65)^{0}.(0.35)^{2} = 0.1225$

n = 3

$P(X = 0) = C_{3,0}.(0.65)^{0}.(0.35)^{3} = 0.0429$

n = 4

$P(X = 0) = C_{4,0}.(0.65)^{0}.(0.35)^{4} = 0.015$

We need 4 stations to be 98% certain that an enemy plane flying over will be detected by at least one station.

(b) If seven stations are in use, what is the expected number of stations that will detect an enemy plane?

The expected number of sucesses of a binomial variable is given by:

$E(x) = np$

So when $n = 7$

$E(x) = 7*(0.65) = 4.55$

If seven stations are in use, the expected number of stations that will detect an enemy plane is 4.55.

6 0

2 years ago

Please help Asap. Will give brainiest if answered correctly with a math question

ivann1987 [24]

TV = VX.

3x-24 = 2x + 1

Add 24 to both sides:

3x = 2x + 25

Subtract 2x from both sides:

X = 24

VX = 2(25) + 1 = 50 + 1 = 51

The answer is 51

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2 years ago

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andrezito [222]

Answer:

The probability that the customer must open 3 or more bottles before finding a prize is 0.64

Step-by-step explanation:

In order for a customer to have to open at least 3 bottles before winning a prize, then the first two bottles shouldnt have a price. The probability that a bottle doesnt have a price is 1-0.2 = 0.8. Since the bottles are independent from each other, then the probability that 2 bottles dont have a prize is 0.8² = 0.64. Therefore, the probability that the customer must open 3 or more bottles before finding a prize is 0.64

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3 years ago

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Need helps asap, im tryna get all of my work done today. So someone answer this fastt

xeze [42]

(4,-5) is Point C

(-5,-4) is point H

(1,-1) is point G

(3,4) is point P

(5,0) is point Z

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