1. A waste management service attempts to design routes so that each of their trucks pick-up on average four tons of garbage or less. A garbage collector believes, however, that he averages picking up more than four tons of garbage per day and decides to perform a hypothesis test. If the hypothesis test is performed at a 5% significance level and the resulting p-value is 0.04. Your conclusion should be: 2. It has been determined with 95% confidence that the proportion of on-line students at NYU who live in Brooklyn is between 0.73 and 0.77. Determine the sample proportion of on-line NYU students who live in Brooklyn.
3. Assume a normal distribution and use a hypothesis test to test the given claim.

According to city reports, it was found that the mean age of the prison population in the city was 26 years. Marc wants to test the claim that the mean age of the prison population in his city is less than 26 years. He obtains a random sample of 25 prisoners, and finds a mean age of 24.4 years and a standard deviation of 9.2 years. At a significance level of 0.05, what should his conclusion be?

1.-Then we p-value indicates that we are in the rejection region we reject H₀

We support the claim of the garbage collector the average picking up more than 4 tn of garbage

2.- p  = 75 %

3.-3.-t(s) is in the rejection region we accept H₀ we have not evidence to support Marc´s claim

Step-by-step explanation:

1.- If p-value is 0,04   and significance level α = 5 %  or  α = 0,05 then p-value < α

Test hypothesis should be    ( x the average of garbage)      

Null hypothesis          H₀           x = 4 Tn

Alternative Hypothesis     Hₐ    x  > 4 Tn

Then  alternative hypothesis suggests a one tail-test to the right  and

p-value <  0,05

Then we p-value indicates that we are in the rejection region we reject H₀

We support the claim of the garbage collector the average picking up more than 4 tn of garbage

2.- As we are dealing with a normal distribution the CI  95 % is symmetrical with respect to the mean, therefore the proportion of student living in Brooklyn is:  

(0,73 + 0,77) /2

p = 0,75     p  = 75 %

Test hypothesis:

Null Hypothesis            H₀            μ    =  26

Alternative Hypothesis      Hₐ      μ   <  26

Alternative hypothesis tells us the test is a one-tail test to the left

Sample size   n = 25

Sample mean   μ  =  24,4

Sample Standard deviation  =  9,2

We assume normal distribution, and as n < 30 we use t-student table

with  24 degree of freedom

Significance level is 0,05  and df = 24  we find t (c)  in t- student table

t(c) = 1,7109      test to the left   t(c) = -1,7109

To calculate  t(s)

t(s)  =  (  24,4 - 26 ) / 9,2 / √25

t(s) =  - 1,6 * 5 / 9,2

t(s) = - 0,87

Comparing t(s)  and t(c) we have

t(s) > t(c)

t(s) is in the rejection region we accept H₀ we have not evidence to support Marc´s claim

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

the revenue of the fitness center, y, is the product of two factors: the number of members and the fee paid by each member. there are 320 members now, but number of members goes down by 10 for each $5 increase in fees, so if x is the number of $5 price increases then members is equal to (320 - 10x). since the fee is currently $45, and the fee increases by $5 for every price bump x, the fee is given by (45 + 5x). so, revenue y is given by y = (320 - 10x)(45 + 5x).