Question

Harry deposits $2000 in a saving account that compounds annually, and plans to leave the money in the account for 2 years. If Harry wants his account to increase in value to $3300, then the interest rate must be greater than what percent?

Answer 1

50% ??

Answer 2

no its multiple choice

Answer 3

so its either 21.45%


16.65%


28.45%


18.65%

Answer 4

its either a or c

Answer 5

how would I solve

Answer 6

???

Answer 7

\(\bf A=P\left(1+\frac{r}{n}\right)^{nt}
\\ \quad \\
A=\textit{current amount}\to 3,300\\
P=\textit{original amount deposited}\to \$2,000\\
{\color{blue}{ r}}=rate\\
n=\textit{times it compounds per year, annually means}\to 1\\
t=years\to 2
\\ \quad \\ \quad \\

3,300=2,000\left(1+\frac{{\color{blue}{r}}}{1}\right)^{1\cdot 2}\)

solve for "r" to find the rate, in decimal format, multiply by 100 to get the % value

Answer 8

ok let me calculate it

Answer 9

18.65%?

Answer 10

hmmm did you solve for "r"?

Answer 11

yea hold on let me redo

Answer 12

i got 13/20 actually

Answer 13

hmmm... so... what's "r" equals to?

Answer 14

hmm umm give me a clue im stuck

Answer 15

:/

Answer 16

is it 120????

Answer 17

well... let us solve for "r" first \( \bf A=P\left(1+\frac{r}{n}\right)^{nt}\implies \cfrac{A}{P}=\left(1+\frac{r}{n}\right)^{nt}
\\ \quad \\
\large\sqrt[nt]{\cfrac{A}{P}}=1+\frac{r}{n}\implies \sqrt[nt]{\cfrac{A}{P}}-1=\cfrac{r}{n}\implies n\left(\sqrt[nt]{\cfrac{A}{P}}-1\right)={\color{blue}{ r}}\)

Answer 18

i dont even understand that

Answer 19

well.. the idea being.... to get "r" or rate

Answer 20

@alyygirl You will have to do better than that. You may need to back up and review some algebra skills.

\(3300 = 2000\cdot(1+r)^{2}\)

Can you solve this for 'r'? Please demonstrate your PROCESS and NOT just your answer.

Answer 21

\(\bf 3,300=2,000\left(1+\frac{{\color{blue}{r}}}{1}\right)^{1\cdot 2}\implies 3,300=2,000(1+{\color{blue}{ r}})^2\) thus

Answer 22

subtract 2000 from both sides so

Answer 23

2000 is attached to the parentheses by multiplication. It takes division to remove it. Divide by 2000.

Answer 24

1.65

Answer 25

You will need to review some algebra.

1.65 is no good. You used to have an equation. Where did it go? Divide both sides by 2000 and have an equation when you are done.

Answer 26

1.65=1(1+r/1) ^1.2?

Answer 27

Feel free to stop working with the silly formula with all the '1's in it.

You had 1*2 in the exponent, Has this now turned into 1.2? I can't tell.

\(3300 = 2000\cdot(1+r)^2\)

Dividing by 2000 gives

\(1.65 = (1+r)^2\)

Now what?

Answer 28

yea 1.2

Answer 29

-2.28

Answer 30

Okay, 1.2 is no good. It was multiplication. \(1\cdot 2 = 2 \ne 1.2\)

Answer 31

Please don't do that. "-2.28" is not helpful. You gave no indication how you managed that.

You have \(1.65 = (1+r)^{2}\)

What is next on your path to a solution. The square seems to be a problem.

Answer 32

divide

Answer 33

?

Answer 34

It's an exponent. You cannot divide it away. It requires a root. The exponent is 2. You need the square root. Go!

Answer 35

how would I get the square root my multiplying?

Answer 36

@alyygirl A little background. Why are you in this class? Did you take a placement exam? What were the prerequisites? How much pressure are you under to pass this course NOW?

You cannot get a square root by multiplying. Do you have a calculator?

Answer 37

Im in a lot of pressure yes I tested into

Answer 38

yes I have a calculator

Answer 39

its my last class

Answer 40

basically my last math thank god

Answer 41

For now, I am not at all impressed by your algebra skills. If I had administered the placement exam, I would have suggested that you spend more time in algebra before taking this course. You will continue to struggle mightily unless you get some significant local help with your algebra. There is a lot more to come and you seem to be missing some very important ideas. Without some serious help, I do not see how you can pass this course. You may need to have a very serious and honest chat with your academic advisor.

You will need to ask your calculator to find the square root of 1.65

\(1.65 = (1+r)^{2}\)

Square Roots

\(1.2845233 = 1+r\)

Finally, subtract one (1) from both sides and you are done.

Answer 42

honestly I dont care because Im a nursing major plus I took my a-g's in highschool just havent done math in a long time.

Answer 43

Caring is one thing. Actually doing the work is another. You WILL need to brush up in order to get through this course. I sincerely hope your course load will allow the time you will need. Get a local tutor for algebra as soon as you can. Otherwise, you're toast.

Good luck.

Answer 44

ok Ill get a tutor but can you atleast work with me on this problem

Answer 45

Ive been passing the other stuff in algebra so far its the word problems

Answer 46

We already solved this one problem. Every step has been shown to you. If you still think you need to work on this problem, that is less encouraging. We are happy to help where we can, but you need a whole algebra course. I fear you don't have time for that.

1) You tried subtraction where multiplication was required.
2) You had no idea when a square root was needed.
3) You found most of the notation baffling.
4) You don't seem to have any reliable way to solve problems even after you have an equation.

I don't think we can blame this on "Word Problems".

I'm just being honest. I want to help you. If you follow my advice, that is the best help I can give you. We can work problems until the cows come home, but without basic skills, we're just not going to get very far. It will be mortally frustrating.

Maybe this one problem will push you over the edge. Do get a local tutor for this one problem. The visit will lead to various conversations and may be all that you need. We can hope. Show the tutor this thread.

Again, I wish you the best of luck.

Answer 47

ok I looked at some steps online

Answer 48

I got 28.45%

Answer 49

Well, since that's the answer I gave you a few posts ago, that will just have to be less impressive than it could have been.

Seriously, consider some local help.