3.2(10)+m(10)=85 this is the correct equation to find how much he walks every morning
The problem ask to explain the use if the normal curve to compare that
score to the population and the best explanation is that the normal
curve depends on a lot of assumptions. If the population score are
normally distributed, we will know its mean and standard deviation, so
can tell that our particular individual's row score falls on that
distributions.
We see that this function has a horizontal asymptote that crosses the y-axis at 2. This means that we can eliminate choices A and C.
From there, we can plug in values for points on the graph into the equation to see if they are true.
f(x)=6(1/3)^x+2
f(1)=6(1/3)^1+2
f(1)=2+2
=4 (Gives us the other coordinate)
f(x)=4(1/3)^x+2
f(1)=4(1/3)^1+2
=(4/3)+2
=3.33 (Does not give us the other coordinate)
Our correct answer choice is B
:)
A) m=months passed
f(m)=2*1.05^m
b) 8 weeks=2 months
f(2)=2*1.05^2=2.205
so essentially still 2
c)100=2*1.05^m
50=1.05^m
log1.05(50)=m
log(50)/log(1.05)=m=80.18
so after 80 months she will have nearly 100 cats (99.12), but will only break through the 100 with her 81st month
Answer:
![a. \ P(X=0)=5.574\times10^{-7}](https://tex.z-dn.net/?f=a.%20%5C%20P%28X%3D0%29%3D5.574%5Ctimes10%5E%7B-7%7D)
![b. \ \ P(X\leq 1)=8.584\times10^{-6}](https://tex.z-dn.net/?f=b.%20%5C%20%5C%20P%28X%5Cleq%201%29%3D8.584%5Ctimes10%5E%7B-6%7D)
![c.\ \ \ P(X\geq 4)=0.9997](https://tex.z-dn.net/?f=c.%5C%20%5C%20%5C%20P%28X%5Cgeq%204%29%3D0.9997)
Step-by-step explanation:
a. #We notice this is a Poisson probability function expressed as:
![f(x)=\frac{\mu^xe^{-\mu}}{x!}](https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7B%5Cmu%5Exe%5E%7B-%5Cmu%7D%7D%7Bx%21%7D)
x-number of occurrences in a given interval.
-mean occurrences of the event
-The mean is calculated as:
![\mu=\frac{14.4}{4}=3.6](https://tex.z-dn.net/?f=%5Cmu%3D%5Cfrac%7B14.4%7D%7B4%7D%3D3.6)
#the probability of no accidents in a 15-minute period is :
![P(x)=\frac{\mu^xe^{-\mu}}{x!}\\\\P(X=0)=\frac{14.4^0e^{-14.4}}{0!}\\\\=5.574\times10^{-7}](https://tex.z-dn.net/?f=P%28x%29%3D%5Cfrac%7B%5Cmu%5Exe%5E%7B-%5Cmu%7D%7D%7Bx%21%7D%5C%5C%5C%5CP%28X%3D0%29%3D%5Cfrac%7B14.4%5E0e%5E%7B-14.4%7D%7D%7B0%21%7D%5C%5C%5C%5C%3D5.574%5Ctimes10%5E%7B-7%7D)
Hence, the probability of no accident in a 15-min period is ![=5.574\times10^{-7}](https://tex.z-dn.net/?f=%3D5.574%5Ctimes10%5E%7B-7%7D)
b. The the probability of at least one accident in a 15-minute period. is calculated as:
![P(x)=\frac{\mu^xe^{-\mu}}{x!}\\\\P(X\leq 1)=\frac{14.4^0e^{-14.4}}{0!}+\frac{14.4^1e^{-14.4}}{1!}\\\\=5.574\times10^{-7}+8.026\times 10^{-6}\\\\=8.584\times 10^{-6}](https://tex.z-dn.net/?f=P%28x%29%3D%5Cfrac%7B%5Cmu%5Exe%5E%7B-%5Cmu%7D%7D%7Bx%21%7D%5C%5C%5C%5CP%28X%5Cleq%201%29%3D%5Cfrac%7B14.4%5E0e%5E%7B-14.4%7D%7D%7B0%21%7D%2B%5Cfrac%7B14.4%5E1e%5E%7B-14.4%7D%7D%7B1%21%7D%5C%5C%5C%5C%3D5.574%5Ctimes10%5E%7B-7%7D%2B8.026%5Ctimes%2010%5E%7B-6%7D%5C%5C%5C%5C%3D8.584%5Ctimes%2010%5E%7B-6%7D)
Hence, the probability of at least one accident in a 15-minute period is ![8.584\times10^{-6}](https://tex.z-dn.net/?f=8.584%5Ctimes10%5E%7B-6%7D)
c. The probability of four or more accidents in a 15-minute period is calculated as:
![P(X\geq 4)=1-P(X\leq 3)=1-[P(X=0)+P(X1)+P(X=2)+P(X=3)]\\\\=1-[5.574\times10^{-7}+a. \ 8.026\times10^{-6}+\frac{14.4^2e^{-14.4}}{2!}+\frac{14.4^3e^{-14.4}}{3!}]\\\\=1-[8.584\times 10^{-6}+5.779\times10^{-5}+2.774\times10^{-4}]\\\\=0.9997](https://tex.z-dn.net/?f=P%28X%5Cgeq%204%29%3D1-P%28X%5Cleq%203%29%3D1-%5BP%28X%3D0%29%2BP%28X1%29%2BP%28X%3D2%29%2BP%28X%3D3%29%5D%5C%5C%5C%5C%3D1-%5B5.574%5Ctimes10%5E%7B-7%7D%2Ba.%20%5C%208.026%5Ctimes10%5E%7B-6%7D%2B%5Cfrac%7B14.4%5E2e%5E%7B-14.4%7D%7D%7B2%21%7D%2B%5Cfrac%7B14.4%5E3e%5E%7B-14.4%7D%7D%7B3%21%7D%5D%5C%5C%5C%5C%3D1-%5B8.584%5Ctimes%2010%5E%7B-6%7D%2B5.779%5Ctimes10%5E%7B-5%7D%2B2.774%5Ctimes10%5E%7B-4%7D%5D%5C%5C%5C%5C%3D0.9997)
Hence,the probability of four or more accidents in a 15-minute period. is 0.9997