$28 x 7 days = $196/week
$196 - $108 = $88
$88 ÷ 0.14 = 628.5 mi
He would need to drive 628.5mi before the Avis price is the same as the Hertz price.
Answer:
![Mean=p=0.75](https://tex.z-dn.net/?f=Mean%3Dp%3D0.75)
![sd=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.75(1-0.75)}{450}}=0.0204](https://tex.z-dn.net/?f=sd%3D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%3D%5Csqrt%7B%5Cfrac%7B0.75%281-0.75%29%7D%7B450%7D%7D%3D0.0204)
Step-by-step explanation:
1) Data given
represent the population proportion of complaints settled for new car dealers
represent the sample of complaints involving new car dealers.
2) Find the distribution of ![\hat p](https://tex.z-dn.net/?f=%5Chat%20p)
First we can begin with the expected value
and that represent the mean
Now we can find the variance for ![\hat p](https://tex.z-dn.net/?f=%5Chat%20p)
When we use a proportion p, when we draw n items each from a Bernoulli distribution. The variance of each Xi distribution is p(1−p) and hence the standard error is p(1−p)/n. for this reason the variance for
is given by:
![Var(\hat p)= \frac{p(1-p)}{n}](https://tex.z-dn.net/?f=Var%28%5Chat%20p%29%3D%20%5Cfrac%7Bp%281-p%29%7D%7Bn%7D)
So then the deviation would be given by:
![Sd(\hat p)=\sqrt{\frac{p(1-p)}{n}}](https://tex.z-dn.net/?f=Sd%28%5Chat%20p%29%3D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D)
The sample distribution of the sample proportion
is normal, so then we have this:
![\hat p \sim N(p,\sqrt{\frac{p(1-p)}{n}})](https://tex.z-dn.net/?f=%5Chat%20p%20%5Csim%20N%28p%2C%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%29)
3) Calculating the mean and standard deviation
We can replace the values given in order to find the mean and deviation:
![Mean=p=0.75](https://tex.z-dn.net/?f=Mean%3Dp%3D0.75)
![sd=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.75(1-0.75)}{450}}=0.0204](https://tex.z-dn.net/?f=sd%3D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%3D%5Csqrt%7B%5Cfrac%7B0.75%281-0.75%29%7D%7B450%7D%7D%3D0.0204)
F(x)=x^3-7x-6 Since I don't have the graph and this is not a perfect cube, I will have to rely on Newton :P
x-(f(x)/(dy/dx))
x-(x^3-7x-6)/(3x^2-7)
(2x^3+6)/(3x^2-7), letting x1=0
0, -6/7, -.988, -.9999, -.99999999999, -1
(x^3-7x-6)/(x+1)
x^2 r -x^2-7x-6
-x r -6x-6
-6 r 0
(x+1)(x^2-x-6)=0
(x+1)(x-3)(x+2)=0
x= -2, -1, 3
How many students would there be? did not see that information if that is the question but if not I say go with the bus that has more seats yes it is more expensive but more space is better!