Calvin deposits $400 in a savings account that accrues 5% interest compounded monthly. after c years, calvin has $658.80. makayl
a deposits $300 in a different savings account that accrues 6% interest compounded quarterly. after m years, makayla has $613.04. what is the approximate difference in the number of years that calvin and makayla have their money invested?
2 answers:
You will need a special formula to compute this.
Years = log (total/principal) / [n * log * (1 + rate / n)]
Part A) Calvin $400 5% monthly 658.80 Time = ?
Monthly compounding "n" = 12
Years = log(658.80/400) / [12 * log(1+ (.05/n))
Years = log ( <span> <span> <span> 1.647 </span> </span> </span> ) / (12 * log ( <span><span><span>1.0041666667 </span> </span> </span> )
Years = 0.21669359917 / 12 * 0.0018058008777
Years = <span> <span> </span></span><span><span>0.21669359917 / 0.0216696105 </span>
</span>Years = <span><span>9.999884362 </span>
</span>
Part B) Makayla 300 6% quarterly 613.04Time=?
Quarterly compounding
n = 4
Years = log (total/principal) / [n * log * (1 + rate / n)]
Years = log (613.04/300) / [4 * log (1 + .06/4)]
Years = log ( <span> <span> <span> 2.0434666667 </span> </span> </span> ) / 4 * log (1.015)
Years = 0.31036755784 / 4 * 0.0064660422492
Years = 0.31036755784 / <span> <span> <span> 0.025864169 </span> </span> </span>
Years = <span> <span> <span> 11.9999044949 </span> </span> </span>
So, the difference is roughly 3 years.
Years = log (total/principal) / [n * log * (1 + rate / n)]
Part A) Calvin $400 5% monthly 658.80 Time = ?
Monthly compounding "n" = 12
Years = log(658.80/400) / [12 * log(1+ (.05/n))
Years = log ( <span> <span> <span> 1.647 </span> </span> </span> ) / (12 * log ( <span><span><span>1.0041666667 </span> </span> </span> )
Years = 0.21669359917 / 12 * 0.0018058008777
Years = <span> <span> </span></span><span><span>0.21669359917 / 0.0216696105 </span>
</span>Years = <span><span>9.999884362 </span>
</span>
Part B) Makayla 300 6% quarterly 613.04Time=?
Quarterly compounding
n = 4
Years = log (total/principal) / [n * log * (1 + rate / n)]
Years = log (613.04/300) / [4 * log (1 + .06/4)]
Years = log ( <span> <span> <span> 2.0434666667 </span> </span> </span> ) / 4 * log (1.015)
Years = 0.31036755784 / 4 * 0.0064660422492
Years = 0.31036755784 / <span> <span> <span> 0.025864169 </span> </span> </span>
Years = <span> <span> <span> 11.9999044949 </span> </span> </span>
So, the difference is roughly 3 years.
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Answer:
103, 776
Step-by-step explanation:
There are 48 possibilities, and you are picking 3.
Now, the order does matter (aka, if the same 3 athletes won but got different places, we would still consider it to be a separate possibility)
(this is considered "without repetitions")
the number of permutations (arrangement combinations where the order does matter) without repetitions formula:
Number of permutations
without <em>repetitions </em>= nPr
=
(
[the P is for combinations [where the order does matter, if the order didn't matter then it would be C, and it would be a different formula.]
n is the total number of objects.
r is the number of objects selected]
<em>(a repetition would be like two versions of the same person competing, which doesn't make sense)</em>
{the ! means factorial:
5! = 5 x 4 x 3 x 2 x 1}
{for example, if there are 5 people to give a presentation, and they can go in any order [but cannot repeat their presentation], so they all must fill 5 slots}
{for the first slot there are 5 choices, the second slot there are 4 choices...}
)
So, if we follow this formula:
n: 48
r: 3
=
=
(the 45! cancel out)
= 48 * 47 * 46
= 103, 776
(or, without the formula:
for the first choice (lets say of gold), there are 48 options
for the second choice (lets say silver), there are 47 options left to choose from
for the third choice (lets say bronze), there are 46 options left to choose from
)