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3 years ago

Please Help!! This is Geometry.

Mathematics

1 answer:

dolphi86 [110]3 years ago

7 0

If M is the midpoint of PQ, then PM and MQ are equal.

As PM and MQ are given , equal them to each other to solve for x.

7x+8=5x+20

Subtract 5x on both sides to get 2x+8=20

Subtract 8 on both sides to get 2x=12

divide by 2 on both sides to get x=6

Use the value of x=6 and plug it into the equations for PM and MQ to find their value.

PM=7x+8

=7(6) 8 =42 + 8 = 50

(you can already tell MQ is also 50 as PM and MQ are equal, but we still solve for MQ just for the fun of it)

MQ=5x + 20

=5(6) +20=30+20=50

PM=50

MQ=50

PQ= PM + MQ= 50 + 50 = 100

Have a great night.

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The probability that your call to a service line is answered in less than 30 seconds is 0.75. Assume that your calls are indepen

vfiekz [6]

Answer:

a) 0.2581

b) 0.4148

c) 17

Step-by-step explanation:

For each call, there are only two possible outcomes. Either they are answered in less than 30 seconds. Or they are not. The probabilities for each call are independent. So we use the binomial probability distribution to solve this question.

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 combinations 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:

$p = 0.75$

a. If you call 12 times, what is the probability that exactly 9 of your calls are answered within 30 seconds? Round your answer to four decimal places (e.g. 98.7654).

This is $P(X = 9)$ when $n = 12$ . So

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

$P(X = 9) = C_{12,9}.(0.75)^{9}.(0.25)^{3} = 0.2581$

b. If you call 20 times, what is the probability that at least 16 calls are answered in less than 30 seconds? Round your answer to four decimal places (e.g. 98.7654).

This is $P(X \geq 16)$ when $n = 20$

$P(X \geq 16) = P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

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

$P(X = 16) = C_{20,16}.(0.75)^{16}.(0.25)^{4} = 0.1897$

$P(X = 17) = C_{20,17}.(0.75)^{17}.(0.25)^{3} = 0.1339$

$P(X = 18) = C_{20,18}.(0.75)^{18}.(0.25)^{2} = 0.0669$

$P(X = 19) = C_{20,19}.(0.75)^{19}.(0.25)^{1} = 0.0211$

$P(X = 20) = C_{20,20}.(0.75)^{20}.(0.25)^{0} = 0.0032$

$P(X \geq 16) = P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.1897 + 0.1339 + 0.0669 + 0.0211 + 0.0032 = 0.4148$

c. If you call 22 times, what is the mean number of calls that are answered in less than 30 seconds? Round your answer to the nearest integer.

The expected value of the binomial distribution is:

$E(X) = np$

In this question, we have $n = 22$

$E(X) = 22*0.75 = 16.5$

The closest integer to 16.5 is 17.

7 0

3 years ago

I need so much help please

jok3333 [9.3K]

Answer:

the Answer will be letter B

6 0

3 years ago

Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one week. Fourteen people ans

Bond [772]

If Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one week. Then the mean of the sample is 3.75.

The average of a group of numbers is simply defined as the mean. In statistics, the mean is regarded as one of the indicators of central tendency.

Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one week.

Fourteen people answered that they generally sell two cars; nineteen generally sell three cars; twelve generally sell four cars; nine generally sell five cars; eleven generally sell six cars.

Then the mean of the sample will be

$\rm Mean = \dfrac{1}{N} \times \Sigma _{i} \ \ x_if_i$