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4 years ago

12

Please answer this question now

1 answer:

soldier1979 [14.2K]4 years ago

7 0

Answer:

6 km

Step-by-step explanation:

You use Pythagoras

${QN}^2 = (PN)^{2}+(PQ)^{2} \\PQ = \sqrt{(QN)^{2}-(PN)^{2} }\\ PQ = 6 km$

You might be interested in

The probability that your call to a service line is answered in less than 30 seconds is 0.85. Assume that your calls are indepen

aev [14]

Answer:

a) 0.1720

b) 0.8298

c) 19

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:

$p = 0.85$

(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

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

$P(X = 9) = C_{12,9}.(0.85)^{9}.(0.15)^{3} = 0.1720$

(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 $n = 20$

So

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

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

$P(X = 16) = C_{20,16}.(0.85)^{16}.(0.15)^{4} = 0.1821$

$P(X = 17) = C_{20,17}.(0.85)^{17}.(0.15)^{3} = 0.2428$

$P(X = 18) = C_{20,18}.(0.85)^{18}.(0.15)^{2} = 0.2293$

$P(X = 19) = C_{20,19}.(0.85)^{19}.(0.15)^{1} = 0.1368$

$P(X = 20) = C_{20,20}.(0.85)^{20}.(0.15)^{0} = 0.0388$

So

$P(X \geq 16) = P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.1821 + 0.2428 + 0.2293 + 0.1368 + 0.0388 = 0.8298$

(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 $n = 22$

So

$E(X) = 22*0.85 = 18.7$

The nearest integer to 18.7 is 19.

7 0

3 years ago

Am I right or if not what is it. :)

dangina [55]

Answer:

Step-by-step explanation:

That is correct.

P=4s=4(12)=48in

A=s^2=12^2=144in^2

8 0

3 years ago

Read 2 more answers

You have $23 in your bank account. You use you debit card and to pay for an item that costs $31. Your bank charges you a $40 ove

vitfil [10]

Your account is gonna be negative 48$

4 0

4 years ago

Read 2 more answers

Suppose we want to build a rectangular storage container with open top whose volume is $$12 cubic meters. Assume that the cost o

Snezhnost [94]

Answer:265.25

Step-by-step explanation:

Given

Volume of box $=12 m^3$

height is 3 times the width

let height, length and breadth be H, L & B

$H=3\times B$

$V=12=L\times B\times H$

4=B^2\times L[/tex]

$L=\frac{4}{B^2}$

Cost of side walls $ $8/m^2$

Cost of base $ $12/m^2$

Cost of side walls $=(2LH+2BH)\cdot 8$

Cost of base $=12LB$

Total cost $C=16LH+16BH+12LB$

$C=48LB+48B^2+12LB$

$C=\frac{192}{B}+48B^2+\frac{48}{B}$

differentiate C w.r.t B to get minimum cost

$\frac{\mathrm{d} C}{\mathrm{d} B}=-\frac{192}{B^2}+2\times 48 B-\frac{48}{B^2}$

$B^3=\frac{240}{2\times 48}$

$B=1.35 m$

$H=3\times 1.35=4.07 m$

L=2.19 m

C=$ 265.25

3 0

3 years ago

What is the slope of a line that contains the points(-13,8) and (-2,-1)

anygoal [31]

Answer:

slope = -(9/11)

Step-by-step explanation:

$slope=\frac{-1-8}{-2-(-13)} =\frac{-9}{-2+13} =-\frac{9}{11}$

I hope this help you

7 0

2 years ago