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alina1380 [7]
3 years ago
6

I will give Brainliest :p

Mathematics
1 answer:
avanturin [10]3 years ago
4 0

Answer:

I think the answer is B

Step-by-step explanation:

Because it's adding 2.5 for every time April Hikes

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the length of a rectangle is 8in. more than it’s width. the perimeter of the rectangle is 24in. what are the width and length of
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The width is 2 in., and the with is 8 in.
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Minimum is 40

First quartile is 43

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Third quartile is 65

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When calculating the effective rate of a loan, which statement or statements must be true if n is greater than 1? I. The length
Fed [463]

When calculating the loan's effective rate, the most accurate statement is that the effective rate will exceed the nominal rate.

<h3>Effective Annual Rate:</h3>

The interest rate for the entire year is known as the effective annual rate (EAR). Interest charges are incurred when a company uses debt or capital leases to fund its operations.

Interest is reported on the income statement, but it can also be generated on an investment or paid on a loan over time due to compounding interest.

It is frequently larger than the marginal rate and is used to compare various financial products with different compounding periods, such as weekly, monthly, and yearly.

The effective yearly interest rate rises over time as the number of compounding periods increases.

Therefore, the correct option is A.

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brainly.com/question/2405320

8 0
3 years ago
Read 2 more answers
Solve the following recurrence relation: <br> <img src="https://tex.z-dn.net/?f=A_%7Bn%7D%3Da_%7Bn-1%7D%2Bn%3B%20a_%7B1%7D%20%3D
-Dominant- [34]

By iteratively substituting, we have

a_n = a_{n-1} + n

a_{n-1} = a_{n-2} + (n - 1) \implies a_n = a_{n-2} + n + (n - 1)

a_{n-2} = a_{n-3} + (n - 2) \implies a_n = a_{n-3} + n + (n - 1) + (n - 2)

and the pattern continues down to the first term a_1=0,

a_n = a_{n - (n - 1)} + n + (n - 1) + (n - 2) + \cdots + (n - (n - 2))

\implies a_n = a_1 + \displaystyle \sum_{k=0}^{n-2} (n - k)

\implies a_n = \displaystyle n \sum_{k=0}^{n-2} 1 - \sum_{k=0}^{n-2} k

Recall the formulas

\displaystyle \sum_{n=1}^N 1 = N

\displaystyle \sum_{n=1}^N n = \frac{N(N+1)}2

It follows that

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

\implies a_n = \dfrac12 n^2 + \dfrac12 n - 1

\implies \boxed{a_n = \dfrac{(n+2)(n-1)}2}

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