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ratelena [41]
3 years ago
12

Kristen invests $ 5745 in a bank. The bank 6.5% interest compounded monthly. How long must she leave the money in the bank for i

t to double? Round to the nearest tenth of a year. Show your work.
How long will it take to triple? Round to nearest tenth of a year. Show your work.
Kristen has a choice to invest her money at 6.5% interest compounded monthly for 5 years or invest her money compounding quarterly at a rate of 6.75% for 5 years. What option would be best for Kristen? Explain and show your work.
Mathematics
2 answers:
AveGali [126]3 years ago
6 0

In order to figure out when the investment will double, we only need to look at the interest rate. The equation to use to find when the investment has doubled is years = ln(2)/ln(1+interest rate), where “ln” is the natural log, and the interest rate is 0.065. Years = ln(2) / ln(1.065) Years = 11.007 = 11.0 years Alternatively, If we want to approximate with the “rule of 72”, (which is a faster, slightly less approximate method), we just divide 72 by the interest rate percentage: Years = 72/6.5 = 11.077 = 11.1 years Alternativley, If we want to approximate with the “rule of 72”, (which is a faster, slightly less approximate method), we just divide 72 by the interest rate percentage: Years = 72/6.5 = 11.077 = 11.1 years





podryga [215]3 years ago
5 0

Answer:

1. 10.7 years

2. 17.0 years

3. 2nd option

Step-by-step explanation:

Use formula for compounded interest

A=P\cdot \left(1+\dfrac{r}{n}\right)^{nt},

where

A is final value, P is initial value, r is interest rate (as decimal), n is number of periods and t is number of years.

In your case,

1. P=$5745, A=2P=$11490, n=12 (compounded monthly), r=0.065 (6.5%) and t is unknown. Then

11490=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12t},\\ \\2=(1.0054)^{12t},\\ \\12t=\log_{1.0054}2,\\ \\t=\dfrac{1}{12}\log_ {1.0054}2\approx 10.7\ years.

2. P=$5745, A=3P=$17235, n=12 (compounded monthly), r=0.065 (6.5%) and t is unknown. Then


17235=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12t},

3=(1.0054)^{12t},

12t=\log_{1.0054}3,

t=\dfrac{1}{12}\log_{1.0054}3\approx 17.0\ years.

3. <u>1 choice:</u> P=$5745, n=12 (compounded monthly), r=0.065 (6.5%), t=5 years and A is unknown. Then

A=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12\cdot 5},\\ \\A=5745\cdot (1.0054)^{60}\approx \$7936.39.

<u>2 choice:</u> P=$5745, n=4 (compounded monthly), r=0.0675 (6.75%), t=5 years and A is unknown. Then

A=5745\cdot \left(1+\dfrac{0.0675}{4}\right)^{4\cdot 5},\\ \\A=5745\cdot (1.0169)^{20}\approx \$8032.58.

The best will be 2nd option, because $8032.58>$7936.39

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3 years ago
Find the value of x in the equation x⁄ 4 + x⁄14 + x⁄17 = 71
Morgarella [4.7K]

Hello from MrBillDoesMath!

Answer:

33769/181

Discussion:

Each term contains "x" so factoring it out gives

x( 1/4 + 1/14 + 1/17)  = 71       (*)

Use common factor (17*14*4  = 952) as the denominator to combine terms:

1/4 =  (17*14)/ 952  = 238/952

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1/17 = (14*4)/952  = 56/952

so 1/4 + 1/14 + 1/17 =    (238 + 68 + 56)/ 952 =  362/952 = 181/476

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4 0
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Consider the functions f and g defined by \[f(x) = \sqrt{\dfrac{x+1}{x-1}}\qquad\qquad\text{and}\qquad\qquad g(x) = \dfrac{\sqrt
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Answer:

The given functions are not same because the domain of both functions are different.

Step-by-step explanation:

The given functions are

f(x)= \sqrt{\dfrac{x+1}{x-1}}

g(x) = \dfrac{\sqrt{x+1}}{\sqrt{x-1}}

First find the domain of both functions. Radicand can not be negative.

Domain of f(x):

\dfrac{x+1}{x-1}>0

This is possible if both numerator or denominator are either positive or negative.

Case 1: Both numerator or denominator are positive.

x+1\geq 0\Rightarrow x\geq -1

x-1\geq 0\Rightarrow x\geq 1

So, the function is defined for x≥1.

Case 2: Both numerator or denominator are negative.

x+1\leq 0\Rightarrow x\leq -1

x-1\leq 0\Rightarrow x\leq 1

So, the function is defined for x≤-1.

From case 1 and 2 the domain of the function f(x) is (-∞,-1]∪[1,∞).

Domain of g(x):

x+1\geq 0\Rightarrow x\geq -1

x-1\geq 0\Rightarrow x\geq 1

So, the function is defined for x≥1.

So, domain of g(x) is [1,∞).

Therefore, the given functions are not same because the domain of both functions are different.

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