Answer:
b because it just is
Step-by-step explanation:
Using the binomial distribution, it is found that there is a 0.8295 = 82.95% probability that at least 5 received a busy signal.
<h3>What is the binomial distribution formula?</h3>
The formula is:
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- 0.54% of the calls receive a busy signal, hence p = 0.0054.
- A sample of 1300 callers is taken, hence n = 1300.
The probability that at least 5 received a busy signal is given by:
![P(X \geq 5) = 1 - P(X < 5)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%205%29%20%3D%201%20-%20P%28X%20%3C%205%29)
In which:
P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).
Then:
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 0) = C_{1300,0}.(0.0054)^{0}.(0.9946)^{1300} = 0.0009](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20C_%7B1300%2C0%7D.%280.0054%29%5E%7B0%7D.%280.9946%29%5E%7B1300%7D%20%3D%200.0009)
![P(X = 1) = C_{1300,1}.(0.0054)^{1}.(0.9946)^{1299} = 0.0062](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20C_%7B1300%2C1%7D.%280.0054%29%5E%7B1%7D.%280.9946%29%5E%7B1299%7D%20%3D%200.0062)
![P(X = 2) = C_{1300,2}.(0.0054)^{2}.(0.9946)^{1298} = 0.0218](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20C_%7B1300%2C2%7D.%280.0054%29%5E%7B2%7D.%280.9946%29%5E%7B1298%7D%20%3D%200.0218)
![P(X = 3) = C_{1300,3}.(0.0054)^{3}.(0.9946)^{1297} = 0.0513](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20C_%7B1300%2C3%7D.%280.0054%29%5E%7B3%7D.%280.9946%29%5E%7B1297%7D%20%3D%200.0513)
![P(X = 4) = C_{1300,4}.(0.0054)^{4}.(0.9946)^{1296} = 0.0903](https://tex.z-dn.net/?f=P%28X%20%3D%204%29%20%3D%20C_%7B1300%2C4%7D.%280.0054%29%5E%7B4%7D.%280.9946%29%5E%7B1296%7D%20%3D%200.0903)
Then:
P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0009 + 0.0062 + 0.0218 + 0.0513 + 0.0903 = 0.1705.
![P(X \geq 5) = 1 - P(X < 5) = 1 - 0.1705 = 0.8295](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%205%29%20%3D%201%20-%20P%28X%20%3C%205%29%20%3D%201%20-%200.1705%20%3D%200.8295)
0.8295 = 82.95% probability that at least 5 received a busy signal.
More can be learned about the binomial distribution at brainly.com/question/24863377
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To get the perimeter of a circle you need to use the formula 2xpix r
so 2x 3.1416 x 2= 12.56
so the third one counting from left to right
-3x+4y=-18
2x-y=7
the answer qould be C.(2,-3)
-3(2)+4(-3)=-18
-6+(-12)=-18
-18=-18
2(2)-(-3)=7
4+3=7
7=7
C.