First thing to do is to illustrate the problem, Since it was mentioned that work was along the way to training, the order is shown in the picture. Mary's home and workplace are nearer compared to her training center. It is also mentioned that the distance between work and home, denoted as x, is 2/3 of the total distance from home to training. The total distance is (x + 2.5). Thus,
x = 2/3(x+2.5)
x = 2/3 x + 5/3
1/3 x = 5/3
x = 5 km
Thus, the distance from home to work is 5 km. This means that Mary has to walk this distance twice to return home to get her shoes. Then, she will travel again the total distance of 5+2.5 = 7.5 km to get to her training center. So,
Total distance = 2(5km) + 7.5 km
Total distance = 17.5 km
Answer:
a

b

Ca
Cb
Explanation:
From the question we are told that
The sample size is n = 100
The upper limit of the 95% confidence interval is b = 47.2 years
The lower limit of the 95% confidence interval is a = 34.5 years
Generally the sample mean is mathematically represented as

=> 
=> 
Generally the margin of error is mathematically represented as

=> 
=> 
Considering question C a
From the question we are told the confidence level is 90% , hence the level of significance is
=>
The sample size is n = 22
Given that the sample size is not sufficient enough i.e
we will make use of the student t distribution table
Generally the degree of freedom is mathematically represented as

=> 
=> 
Generally from the student t distribution table the critical value of
at a degree of freedom of 21 is
Considering question C b
From the question we are told the confidence level is 80% , hence the level of significance is
=>
Generally from the normal distribution table the critical value of
is
They would be 0.00005524 miles apart because 1 in equals 0.00001578 miles
6^2 = 36
thats what you meant?
7 x 7 = 49. 49/2 = 24.5
The area is 24.5 units squared.