B just trust me bro I got you
P(x)=R(x)-C(x)
=(-0.5x²+800x-100)-(300x+250)
=-0.5x²+800x-100-300x-250
=-0.5x²+800x-300x-100-250
=-0.5x²+500x-350 (2)
It would be 3/19. Cause you add all and that would be denom.
Solution :
To claim to be tested is whether "the mean salary is higher than 48,734".
i.e. μ > 48,734
Therefore the null and the alternative hypothesis are

and 
Here, n = 50

s = 3600
We take , α = 0.05
The test statistics t is given by


t = 2.15
Now the ">" sign in the
sign indicates that the right tailed test
Now degree of freedom, df = n - 1
= 50 - 1
= 49
Therefore, the p value = 0.02
The observed p value is less than α = 0.05, therefore we reject
. Hence the mean salary that the accounting graduates are offered from the university is more than the average salary of 48,734 dollar.
Answer:
h=0.6m
r=0.75m
h=0.6m/0.75m*100 percent of r.
h=6000/75% of r
h=80% of r
Step-by-step explanation: