The decimal for this would be 7.575
Hope this helped :)
Y=mx
m=amount of time per day studied
y=time
x=number of days
we have
60=m(4)
divide by 4
15=m
y=15x
Answer:
The difference in the sample proportions is not statistically significant at 0.05 significance level.
Step-by-step explanation:
Significance level is missing, it is α=0.05
Let p(public) be the proportion of alumni of the public university who attended at least one class reunion
p(private) be the proportion of alumni of the private university who attended at least one class reunion
Hypotheses are:
: p(public) = p(private)
: p(public) ≠ p(private)
The formula for the test statistic is given as:
z=
where
- p1 is the sample proportion of public university students who attended at least one class reunion (
)
- p2 is the sample proportion of private university students who attended at least one class reunion (
)
- p is the pool proportion of p1 and p2 (
)
- n1 is the sample size of the alumni from public university (1311)
- n2 is the sample size of the students from private university (1038)
Then z=
=-0.207
Since p-value of the test statistic is 0.836>0.05 we fail to reject the null hypothesis.
We have that
y=6^x
we know
to find the inverse
replace x with y and y with x,
x=6^y
we know that
y=logb(x)----------> x=b^y
therefore
x=6^y---------------> y=log6(x)
the answer is the option A.) Y=log 6x