has to be differentiable on [-1, 1] for the theorem to hold. This is not the case because the derivative doesn't exist at
.
but clearly
is undefined.
Answer:
a) monthly payment: $177.50
b) total amount paid: $31,950
c) toward principal: $17,500; toward interest: $14,450
Step-by-step explanation:
a) The amount of the monthly payment (A) is computed from the principal (P), the annual interest rate (r) and the number of years (n) using the formula ...
A = P·(r/12)/(1 -(1 +r/12)^(-12n))
Filling in your numbers, we can use r/12 = 0.09/12 = 0.0075, and 12n = 12·15 = 180:
A = $17500·0.0075/(1 - 1.0075^-180) ≈ $177.50
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b) The total payment over the term of the loan is 180 of these monthly payments:
180·$177.50 = $31,950
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c) $17,500 is paid toward the principal.
$14,450 is paid toward interest.
I'm going to rewrite f(t) to make it a little easier to use the power rule:
Now just take the derivative using the power rule:
Since they ask for f'(a), we need to replace the t's with a's:
Center : Mean Before the introduction of the new course, center = average(121,134,106,93,149,130,119,128) = 122.5 After the introduction of the new course, center = average(121,134,106,93,149,130,119,128,45) = 113.9 The center has moved to the left (if plotted in a graph) because of the low intake for the new course. Spread before introduction of the new course : Arrange the numbers in ascending order: (93, 106,119, 121), (128, 130,134, 149) Q1=median(93,106,119,121) = 112.5 Q3=median(128,130,134,149) = 132 Spread = Interquartile range = Q3-Q1 = 19.5 After addition of the new course,
(45,93, 106,119,) 121, (128, 130,134, 149)
Q1=median(45,93,106,119)=99.5
Q3=median (128, 130,134, 149)= 132
Spread = Interquartile range = 132-99.5 =32.5
We see that the spread has increased after the addition of the new course.
