Answer:
f + g = ![\frac{x-1}{x+ 3}+\frac{x-3}{x+1}=\frac{x^{2}-1 }{(x+3)(x+1)}+\frac{x^{2}-9 }{(x+1)(x+3)}=\frac{2x^{2}-10 }{(x+1)(x+3)}](https://tex.z-dn.net/?f=%5Cfrac%7Bx-1%7D%7Bx%2B%203%7D%2B%5Cfrac%7Bx-3%7D%7Bx%2B1%7D%3D%5Cfrac%7Bx%5E%7B2%7D-1%20%7D%7B%28x%2B3%29%28x%2B1%29%7D%2B%5Cfrac%7Bx%5E%7B2%7D-9%20%7D%7B%28x%2B1%29%28x%2B3%29%7D%3D%5Cfrac%7B2x%5E%7B2%7D-10%20%7D%7B%28x%2B1%29%28x%2B3%29%7D)
Step-by-step explanation:
Step-by-step explanation
n^3-n.
Hope this will help you :) <3
One you can make out of paper, the other is paper. ^-^
Answer: the answer is C.
Step-by-step explanation:
Answer:
0.9155
Step-by-step explanation:
As, the random variable X the number of accident follows the Poisson distribution so,
P(X=x)=(μ^x)(e^-μ)/x!
μ=average accident=4.1.
The probability that more than one accident occurs per year= P(X>1)=?
P(X>1)=1-P(X≤1)
P(X>1)=1-[P(X=0)+P(X=1)]
P(X=0)=(4.1^0)(e^-4.1)/0!
P(X=0)=0.016573/1
P(X=0)=0.016573
P(X=1)=(4.1^1)(e^-4.1)/1!
P(X=1)=(4.1)0.016573/1
P(X=1)=0.0679481
P(X>1)=1-[0.016573+0.0679481]
P(X>1)=1-0.084521
P(X>1)=0.915479
P(X>1)=0.9155
The probability that more than one accident occurs per year is 91.55%