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jarptica [38.1K]
4 years ago
13

Amount Financed (m) = $2,500 Number of Payments per year (y) = 12 Number of Payments (n) = 36 Total Interest (c) = $158.15

Mathematics
2 answers:
iVinArrow [24]4 years ago
6 0

4.1% is the answer hope i helped

olga55 [171]4 years ago
5 0
The number of payments per year is 12
The number of payments is 36 which means that the term lasted for 3 years.
The total interest is $158.15 and the amount financed is $2500.
The interest rate is
158.15/2500 = 6.33%
In nominal month, that is
(1 + r/12)^12 - 1 = 0.0633
Solve for r
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Answer:

Less than high school” and “ high school graduate”

Step-by-step explanation:

I know.

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A tree casts a shadow of 24 ft. At the same time, a 6 ft adult casts a shadow of 2 ft. what is the hight of the tree?
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Answer: 8ft tall

Step-by-step explanation:

6/2=3

24/3=8

so the tree is 8ft tall

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1. A clerk at the pet store kept track of the number of fish purchased by customers during one week. Mary bought 1 fish, Ari bou
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3 years ago
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Evaluate the Riemann sum for f(x) = 3 - 1/2 times x between 2 and 14 where the endpoints are included with six subintervals taki
Digiron [165]

Answer:

-6

Step-by-step explanation:

Given that :

we are to evaluate the Riemann sum for f(x) = 3 - \dfrac{1}{2}x from 2 ≤ x ≤ 14

where the endpoints are included with six subintervals, taking the sample points to be the left endpoints.

The Riemann sum can be computed as follows:

L_6 = \int ^{14}_{2}3- \dfrac{1}{2}x \dx = \lim_{n \to \infty} \sum \limits ^6 _{i=1} \ f (x_i -1) \Delta x

where:

\Delta x = \dfrac{b-a}{a}

a = 2

b =14

n = 6

∴

\Delta x = \dfrac{14-2}{6}

\Delta x = \dfrac{12}{6}

\Delta x =2

Hence;

x_0 = 2 \\ \\  x_1 = 2+2 =4\\ \\  x_2 = 2 + 2(2) \\ \\  x_i = 2 + 2i

Here, we are  using left end-points, then:

x_i-1 = 2+ 2(i-1)

Replacing it into Riemann equation;

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} \begin {pmatrix}3 - \dfrac{1}{2} \begin {pmatrix}  2+2 (i-1)  \end {pmatrix} \end {pmatrix}2

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 6 - (2+2(i-1))

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 6 - (2+2i-2)

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 6 -2i

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 6 -   \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 2i

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 6 - 2  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} i

Estimating the integrals, we have :

= 6n - 2 ( \dfrac{n(n-1)}{2})

= 6n - n(n+1)

replacing thevalue of n = 6 (i.e the sub interval number), we have:

= 6(6) - 6(6+1)

= 36 - 36 -6

= -6

5 0
3 years ago
In a sequence of numbers, a5 = 30, a6 = 34, a7 = 38, a8 = 42, and a9 = 46.
defon

Answer:

A.)

Step-by-step explanation:

Try each choice to see which one works.

A.)

a_n = 4n + 10

a_5 = 4(5) + 10 = 30

a_6 = 4(6) + 10 = 34

a_7 = 4(7) + 10 = 38

Answer: A.)

7 0
2 years ago
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