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Lisa [10]
3 years ago
15

1,280 at 3% compounded annually for 3 years

Mathematics
1 answer:
ladessa [460]3 years ago
3 0

Answer:42,660

Step-by-step explanation:

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Can anyone one give me the answer for this??!! Please hurry up to
Marat540 [252]

Answer:

N=6

Step-by-step explanation:

As the answer is 6 and we already have 3 so the other half is 3. Half of 6 is three so n=6

5 0
3 years ago
Suppose that $2000 is invested at a rate of 2.6% , compounded semiannually. Assuming that no withdrawals are made, find the tota
EleoNora [17]

Answer:

$2,589.52

Step-by-step explanation:

A = P(1 + \dfrac{r}{n})^{nt}

We start with the compound interest formula above, where

A = future value

P = principal amount invested

r = annual rate of interest written as a decimal

n = number of times interest is compound per year

t = number of years

For this problem, we have

P = 2000

r = 0.026

n = 2

t = 10,

and we find A.

A = $2000(1 + \dfrac{0.026}{2})^{2 \times 10}

A = $2589.52

8 0
3 years ago
Read 2 more answers
(3x+2)+(2x^2+3x-1)<br><br> Please answer ty:)
Paul [167]
Answer = 2x^2+6x+1

3x+2+2x^2+3x-1
3x+1+2x^2+3x
Now combine like terms
3x+1+2x^2+3x
6x+1+2x^2
Rearrange terms
6x+1+2x^2
2x^2+6x+1
4 0
2 years ago
A candy store has 6 boxes of chocolates. Each box has 500 pieces How many pieces are there altogether in the 6 boxes?​
arsen [322]
Answer is 3,000 hope it helps
6 0
2 years ago
having trouble with this problem, pls help :/ I know the answer, just not how to get there. (ap calc ab, integrals)
pochemuha

Answer: D) 101

Step-by-step explanation:

By linearity, we can break it up into 2 integrals. The integral and derivative of f easily cancel out

\int\limits^{10}_{-1} {(2x+0.5f'(x))} \, dx =\int\limits^{10}_{-1} {2x} \, dx +0.5\int\limits^{10}_{-1} {f'(x)} \, dx =x^2|^{^{10}}_{_{-1}}+0.5f(x)|^{^{10}}_{_{-1}}\\=(100-1)+0.5(f(10)-f(-1))=99+0.5(8-4))=101

I used the table for values of f(x) at 10 and -1. Wouldn't be surprised if this was part of a series of questions about f because I really can't see how you could use the hypothesis that f is twice differentiable on R. Same for the other table values. I'm curious about how you found the answer. Was it a different way?

7 0
2 years ago
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