How many lines of symmetry does a object shaped like a stop sign have

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

3 years ago

Are your finances, buying habits, medical records, and phone calls really private? A real concern for many adults is that comput

Andrej [43]

Answer:

a)There is a 4.88% probability that none is concerned that employers are monitoring phone calls.

b)There is a 7.89% probability that all are concerned that employers are monitoring phone calls.

c)There is a 37.23% probability that exactly two are concerned that employers are monitoring phone calls.

Step-by-step explanation:

The binomial probability is the probability of exactly x successes on n repeated trials in an experiment which has two possible outcomes (commonly called a binomial experiment).

It is given by the following formula:

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

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 a success.

In this problem, a success is being concerned that employers are monitoring phone calls.

53% of adults are concerned that employers are monitoring phone calls, so $p = 0.53$

(a) Out of four adults, none is concerned that employers are monitoring phone calls.

Four adults, so $n = 4$ .

Is the probability of 0 successes, so x = 0.

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

$P = C_{4,0}.(0.53)^{0}.(0.47)^{4}$

$P = 0.0488$

There is a 4.88% probability that none is concerned that employers are monitoring phone calls.

(b) Out of four adults, all are concerned that employers are monitoring phone calls.

Four adults, so $n = 4$ .

Is the probability of 4 successes, so x = 4.

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

$P = C_{4,0}.(0.53)^{4}.(0.47)^{0}$

$P = 0.0789$

There is a 7.89% probability that all are concerned that employers are monitoring phone calls.

(c) Out of four adults, exactly two are concerned that employers are monitoring phone calls.

Four adults, so $n = 4$ .

Is the probability of 4 successes, so x = 2.

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

$P = C_{4,2}.(0.53)^{2}.(0.47)^{2}$

$P = 0.3723$

There is a 37.23% probability that exactly two are concerned that employers are monitoring phone calls.

3 0

3 years ago

The length of a rectangle is (x + 2). The

Oksanka [162]

Answer:

a

Step-by-step explanation:

7 0

3 years ago

Alice has 1/5 as many miniature cars as Sylvester has Sylvester has 35 miniature cars how many miniature cars does alic have?

const2013 [10]

The answer is 7. Since Alice has 1/5 as many cars as Sylvester. You multiply 1/5 by how many cars Sylvester has. Sylvester has 35 cars, so you would do 35 times 1/5 or 35 divided by 5 (multiply by 1/5 and divided by 5 are the same thing) and you would get your answer of 7.

6 0

3 years ago

Can someone help me with these 2 problems plz

Taya2010 [7]

For the first one I believe that the numbers are both odd and I am not sure about the second one sorry

4 0

3 years ago

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