anonymous
  • anonymous
You decide to put $100 in a savings account to save for a $3,000 down payment on a new car. If the account has an interest rate of 2% per year and is compounded monthly, how long does it take you to earn $3,000 without depositing any additional funds?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@ospreytriple
anonymous
  • anonymous
@Mertsj
anonymous
  • anonymous
Do you know the compound interest formula from your course?

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More answers

anonymous
  • anonymous
yeah
anonymous
  • anonymous
a(t)=p(1+r)^t
anonymous
  • anonymous
OK. It looks like we want a(t) to be $3000 cause that's how much needs to be saved. p is the present value. How much is that?
anonymous
  • anonymous
100
anonymous
  • anonymous
OH!! In your previous question, the right hand side of the equation was NEGATIVE, so the correct answer is the negative number. Sorry about that.
anonymous
  • anonymous
$100 is correct.
anonymous
  • anonymous
The interest rate (r) is a little bit tricky. The question says the nominal yearly interest rate is 2% but it is compounded monthly. Do you know how to get the interest rate in this case?
anonymous
  • anonymous
ummm 12??
anonymous
  • anonymous
Well, it's 2% per year, but for a month it's going to be 1/12 as much. What do you think?
anonymous
  • anonymous
I agree
anonymous
  • anonymous
OK. so r = 2% /12. Have to convert that to a decimal. What do you get?
anonymous
  • anonymous
0.02
anonymous
  • anonymous
Right. But you still have to divide by 12 to get r.
anonymous
  • anonymous
ok so 0.02/12 ??
Mertsj
  • Mertsj
\[A=P(1+\frac{r}{n})^{nt}\] \[3000=100(1+\frac{.02}{12})^{12t}\]
Mertsj
  • Mertsj
Solve for t
anonymous
  • anonymous
That's it. What's r then?
anonymous
  • anonymous
3000=100.16667)^12t
anonymous
  • anonymous
r is not 0.16667. Try it again
anonymous
  • anonymous
r=0.02
anonymous
  • anonymous
Remember, we already established that r = 0.02/12. What do you get?
anonymous
  • anonymous
0.001666667
anonymous
  • anonymous
Great. Now out that value of r into the compound interest formula and what do you get?
Mertsj
  • Mertsj
r=.02
anonymous
  • anonymous
Perhaps you'd like to take over @Mertsj
anonymous
  • anonymous
3000=(100.16667)^12t
anonymous
  • anonymous
What's 100 + 0.0016667 ?
Mertsj
  • Mertsj
Continuing: \[\frac{3000}{100}=\frac{100(1+\frac{.02}{12})^{12t}}{100}\]
anonymous
  • anonymous
100.0016667
anonymous
  • anonymous
30=(1+0.02/12)^12t
anonymous
  • anonymous
Great. So you have\[3000 = 100.0016667^{12t}\]Now take the log of both sides
Mertsj
  • Mertsj
\[300=(1.01666666)^{12t}\]
anonymous
  • anonymous
ok so 1og 300=....?
Mertsj
  • Mertsj
\[\log_{10}300=12t \log_{10}1.0166666666 \]
anonymous
  • anonymous
I'm sorry @BellaBlue77 . Our work is being hijacked. I'm out. Good luck.
Mertsj
  • Mertsj
\[\frac{\log_{10}300}{\log_{10}1.016666666}=12t \]
anonymous
  • anonymous
ok hold onn
anonymous
  • anonymous
300/ 1.01666666=12t
Mertsj
  • Mertsj
\[345.07=12t\]
anonymous
  • anonymous
28. 76
Mertsj
  • Mertsj
\[28.8=t\]
anonymous
  • anonymous
170.202 years 14.3129 years 171.755 years 168.354 years
anonymous
  • anonymous
That's not an answer choice
anonymous
  • anonymous
@heretohelpalways @ganeshie8
Mertsj
  • Mertsj
Error: 3000/100=30
anonymous
  • anonymous
yes
Mertsj
  • Mertsj
\[\frac{\log_{10}30}{\log_{10}1.0166666666}=12t \]
anonymous
  • anonymous
ok so 30/1.016666
anonymous
  • anonymous
29.5081967232=12t
Mertsj
  • Mertsj
you have to take the log of those numbers.
anonymous
  • anonymous
so log 29.508...?
Mertsj
  • Mertsj
Do you see that it says log 30?
anonymous
  • anonymous
1.46994267012
anonymous
  • anonymous
1.47712125472 i mean
anonymous
  • anonymous
would 168.354 years be the answer?
Mertsj
  • Mertsj
I keep getting 17.14
anonymous
  • anonymous
I know :( But thats not an answer choice. 170.202 years 14.3129 years 171.755 years 168.354 years see?
anonymous
  • anonymous
Oh well, let's call it a day loll
anonymous
  • anonymous
If you find the answer, message me pleae, I'll be back tommorw
Mertsj
  • Mertsj
Starting over: \[3000=100(1+\frac{.02}{12})^{12t}\] \[30=(1+.00166666666)^{12t}\]
Mertsj
  • Mertsj
\[\frac{\log_{10}30}{\log_{10}1.00166666666666}=12t \]
Mertsj
  • Mertsj
\[2042.4185=12t\]
Mertsj
  • Mertsj
\[170.20=t\]

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