Answer:
0.9855 or 98.55%.
Step-by-step explanation:
The probability of each individual match being flawed is p = 0.008. The probability that a matchbox will have one or fewer matches with a flaw is the same as the probability of a matchbox having exactly one or exactly zero matches with a flaw:
![P(X\leq 1)=P(X=0)+P(X=1)\\P(X\leq 1)=(1-p)^{23}+23*(1-p)^{23-1}*p\\P(X\leq 1)=(1-0.008)^{23}+23*(1-0.008)^{23-1}*0.008\\P(X\leq 1)=0.8313+0.1542\\P(X\leq 1)=0.9855](https://tex.z-dn.net/?f=P%28X%5Cleq%201%29%3DP%28X%3D0%29%2BP%28X%3D1%29%5C%5CP%28X%5Cleq%201%29%3D%281-p%29%5E%7B23%7D%2B23%2A%281-p%29%5E%7B23-1%7D%2Ap%5C%5CP%28X%5Cleq%201%29%3D%281-0.008%29%5E%7B23%7D%2B23%2A%281-0.008%29%5E%7B23-1%7D%2A0.008%5C%5CP%28X%5Cleq%201%29%3D0.8313%2B0.1542%5C%5CP%28X%5Cleq%201%29%3D0.9855)
The probability that a matchbox will have one or fewer matches with a flaw is 0.9855 or 98.55%.
B is the correct answer R=Q-7s/4
Using the Pythagorean theorem we can solve for the radius
see attached picture:
image ????? please attach image and try again
-3/5
-6/10
-.06 is the answer