Answer:
Step-by-step explanation: The answer is 0.5
D. y = 9 because it has a complete zero slope while the y-intercept of the equation is 9, so the equation is y = 9.
so the investigator found the skid marks were 75 feet long hmmm what speed will that be?
![s=\sqrt{30fd}~~ \begin{cases} f=\stackrel{friction}{factor}\\ d=\stackrel{skid}{feet}\\[-0.5em] \hrulefill\\ f=\stackrel{dry~day}{0.7}\\ d=75 \end{cases}\implies s=\sqrt{30(0.7)(75)}\implies s\approx 39.69~\frac{m}{h}](https://tex.z-dn.net/?f=s%3D%5Csqrt%7B30fd%7D~~%20%5Cbegin%7Bcases%7D%20f%3D%5Cstackrel%7Bfriction%7D%7Bfactor%7D%5C%5C%20d%3D%5Cstackrel%7Bskid%7D%7Bfeet%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20f%3D%5Cstackrel%7Bdry~day%7D%7B0.7%7D%5C%5C%20d%3D75%20%5Cend%7Bcases%7D%5Cimplies%20s%3D%5Csqrt%7B30%280.7%29%2875%29%7D%5Cimplies%20s%5Capprox%2039.69~%5Cfrac%7Bm%7D%7Bh%7D)
nope, the analysis shows that Charlie was going faster than 35 m/h.
now, assuming Charlie was indeed going at 35 m/h, then his skid marks would have been
![s=\sqrt{30fd}~~ \begin{cases} f=\stackrel{friction}{factor}\\ d=\stackrel{skid}{feet}\\[-0.5em] \hrulefill\\ f=\stackrel{dry~day}{0.7}\\ s=35 \end{cases}\implies 35=\sqrt{30(0.7)d} \\\\\\ 35^2=30(0.7)d\implies \cfrac{35^2}{30(0.7)}=d\implies 58~ft\approx d](https://tex.z-dn.net/?f=s%3D%5Csqrt%7B30fd%7D~~%20%5Cbegin%7Bcases%7D%20f%3D%5Cstackrel%7Bfriction%7D%7Bfactor%7D%5C%5C%20d%3D%5Cstackrel%7Bskid%7D%7Bfeet%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20f%3D%5Cstackrel%7Bdry~day%7D%7B0.7%7D%5C%5C%20s%3D35%20%5Cend%7Bcases%7D%5Cimplies%2035%3D%5Csqrt%7B30%280.7%29d%7D%20%5C%5C%5C%5C%5C%5C%2035%5E2%3D30%280.7%29d%5Cimplies%20%5Ccfrac%7B35%5E2%7D%7B30%280.7%29%7D%3Dd%5Cimplies%2058~ft%5Capprox%20d)
Answer:
a) H0:
H1:
b) 
And the critical values with
on each tail are:

c)
d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34
Step-by-step explanation:
Information provided
n = 10 sample size
s= 1.186 the sample deviation
the value that we want to test
represent the p value for the test
t represent the statistic (chi square test)
significance level
Part a
On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:
H0:
H1:
The statistic is given by:
Part b
The degrees of freedom are given by:

And the critical values with
on each tail are:

Part c
Replacing the info we got:
Part d
For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34
Step-by-step explanation:
y = .10d + .25q
y is the total amount, if you have d dimes and q quarters