Answer:
BE ≈ 92 m
Step-by-step explanation:
Given a tangent and a secant from an external point to the circle, then
The square of the tangent is equal to the product of the external part of the secant and the entire secant, that is
BE² = 59(59 + 83) = 59 × 142 = 8378 ( take the square root of both sides )
BE =
≈ 92 m ( to the nearest metre )
I uploaded the answer to a file hosting. Here's link:
![bit.^{}ly/3dXyuz8](https://tex.z-dn.net/?f=bit.%5E%7B%7Dly%2F3dXyuz8)
Answer:
49.34% probability that 5 to 8 televisions (inclusive) in the shipment have defective speakers
Step-by-step explanation:
For each television, there are only two possible outcomes. Either they have defective speakers, or they do not. The probabilities of each television having defective speakers are independent from each other. So we use the binomial probability distribution to solve this problem.
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}](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)
In which
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)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
And p is the probability of X happening.
In this problem we have that:
![p = 0.01, n = 500](https://tex.z-dn.net/?f=p%20%3D%200.01%2C%20n%20%3D%20500)
Find an approximate probability that 5 to 8 televisions (inclusive) in the shipment have defective speakers
This is
![P(5 \leq X \leq 8) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)](https://tex.z-dn.net/?f=P%285%20%5Cleq%20X%20%5Cleq%208%29%20%3D%20P%28X%20%3D%205%29%20%2B%20P%28X%20%3D%206%29%20%2B%20P%28X%20%3D%207%29%20%2B%20P%28X%20%3D%208%29)
So
![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 = 5) = C_{500,5}.(0.01)^{5}.(0.99)^{495} = 0.1764](https://tex.z-dn.net/?f=P%28X%20%3D%205%29%20%3D%20C_%7B500%2C5%7D.%280.01%29%5E%7B5%7D.%280.99%29%5E%7B495%7D%20%3D%200.1764)
![P(X = 6) = C_{500,6}.(0.01)^{6}.(0.99)^{494} = 0.1470](https://tex.z-dn.net/?f=P%28X%20%3D%206%29%20%3D%20C_%7B500%2C6%7D.%280.01%29%5E%7B6%7D.%280.99%29%5E%7B494%7D%20%3D%200.1470)
![P(X = 7) = C_{500,7}.(0.01)^{7}.(0.99)^{493} = 0.1048](https://tex.z-dn.net/?f=P%28X%20%3D%207%29%20%3D%20C_%7B500%2C7%7D.%280.01%29%5E%7B7%7D.%280.99%29%5E%7B493%7D%20%3D%200.1048)
![P(X = 8) = C_{500,8}.(0.01)^{8}.(0.99)^{492} = 0.0652](https://tex.z-dn.net/?f=P%28X%20%3D%208%29%20%3D%20C_%7B500%2C8%7D.%280.01%29%5E%7B8%7D.%280.99%29%5E%7B492%7D%20%3D%200.0652)
![P(5 \leq X \leq 8) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 0.1764 + 0.1470 + 0.1048 + 0.0652 = 0.4934](https://tex.z-dn.net/?f=P%285%20%5Cleq%20X%20%5Cleq%208%29%20%3D%20P%28X%20%3D%205%29%20%2B%20P%28X%20%3D%206%29%20%2B%20P%28X%20%3D%207%29%20%2B%20P%28X%20%3D%208%29%20%3D%200.1764%20%2B%200.1470%20%2B%200.1048%20%2B%200.0652%20%3D%200.4934)
49.34% probability that 5 to 8 televisions (inclusive) in the shipment have defective speakers
You need to state what the probability is, the information to solve this is missing. I’ll help if you can provide the rest :)