Answer:
15.14%
Step-by-step explanation:
The formula for APR is stated thus:
APR=fees+interest/principal/n*365*100
principal is the loan amount of $700
fees is the processing fees on the loan which is $50
interest amount=principal*interest %=$700*8%=$56
n is the number of days of the loan which is a year i.e 365 days
APR=($50+$56)/$700/365*365*100
APR=$106/$700/365*365*100
APR=0.151428571
/365*365*100
APR=0.151428571
*100=15.14%
The annual percentage rate on the loan is 15.14% which represents the actual cost on the loan not just the interest cost of 8% annually
 
        
             
        
        
        
I'd be more than happy to help. Im a sophmore and Love math
        
                    
             
        
        
        
90.42857143 or
90.4 to nearest decimal
        
             
        
        
        
Answer:
a. We reject the null hypothesis at the significance level of 0.05
b. The p-value is zero for practical applications
c. (-0.0225, -0.0375)
Step-by-step explanation:
Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.  
Then we have  ,
,  ,
,  and
 and  ,
,  ,
,  . The pooled estimate is given by
. The pooled estimate is given by  
 
a. We want to test  vs
 vs  (two-tailed alternative).
 (two-tailed alternative).  
The test statistic is  and the observed value is
 and the observed value is  . T has a Student's t distribution with 20 + 25 - 2 = 43 df.
. T has a Student's t distribution with 20 + 25 - 2 = 43 df.
The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value  falls inside RR, we reject the null hypothesis at the significance level of 0.05
 falls inside RR, we reject the null hypothesis at the significance level of 0.05
b. The p-value for this test is given by  0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.
0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.
c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)
 , i.e.,
, i.e.,
 
where  is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So
 is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So
 , i.e.,
, i.e.,
(-0.0225, -0.0375)
 
        
             
        
        
        
1. Points F and C 
2. 2 
3. (3,4)
4. (1,-2)
5. Quadrant 4
6. (-3,-3) 
(I also added a photo of a graph with the labeled quadrants to help you)