Mathematics, 04.04.2020 07:28 kendra95

A person invests 9000 dollars in a bank. The bank pays 4.25% interest compounded daily. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 19600?

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Answer from: rousedem28381

It will take approximately 18.4years for the person to make 19600

The formula for calculating the compound amount is expressed as:

$A=P(1+\frac{r}{n} )^{nt}$ where:

A is the amount

P is the amount invested

n is the compounding amount

t is the time taken

Given the following parameters

A = 19600

P = 9000

r = 4.25% = 0.0425

n = 365 (compounded daily)

Substitute the given parameters into the formula to get the time "t"

$19600 = 9000(1+\frac{0.0425}{365} )^{365t}\\2.1778 = (1+0.0001164)^{365t}\\2.1778 = (1.0001164)^{365t}\\log2.1778 = 365tlog(1.0001164)\\0.3380 = 0.0184t\\t = \frac{0.3380}{0.0184} \\t \approx 18yrs$

Hence it will take approximately 18.4years for the person to make 19600

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Answer from: Gearyjames8

The person must invest the money in the bank for 18 years to reach $19600, if a person invests 9000 dollars in a bank and the bank pays 4.25% interest compounded daily.

Step-by-step explanation:

The given is,

A person invests 9000 dollars in a bank

The bank pays 4.25% interest compounded daily

Future amount $19600

Step:1

Formula to calculate the future investment with compounded daily,

$A = P (1+\frac{r}{n})^{nt}$ ................................(1)

Where, A - Future amount

P - Initial investment

r - Interested rate

n - No.of times for a year

t - No.of years

Step:2

From the given values,

A = $ 19600

P = $ 9000

r = 4.25 %

n = 365 ( ∵ Compounded daily )

Equation (1) becomes,

$19600 = 9000(1+\frac{0.0425}{365})^{365t}$

$\frac{19600}{9000}= (1+0.000116438)^{365t}$

$2.177778=(1.000116438)^{365t}$

( Take log on both sides )

$log 2.17778= log (1.000116438)^{365t}$

log 2.17778=( 365t)log (1.000116438)

Substitute the log values,

0.338014 = (365t)(0.000050565)

$\frac{0.338014}{0.000050565}= 365t$

6684.6842 = 365t

$t = \frac{6684.6842}{365}$

= 18.314

≅ 18 years

No.of years, t = 18 years

Result:

The person must invest the money in the bank for 18 years to reach $19600, if a person invests 9000 dollars in a bank and the bank pays 4.25% interest compounded daily.

Answer from: Quest

there are many solutions.

step-by-step explanation:

step 1: simplify both sides of the equation.

4x+6=4x^2+10x−4

step 2: subtract 4x^2+10x-4 from both sides.

4x+6−(4x^2+10x−4)=4x^2+10x−4−(4x^2+10x−4)

−4x^2−6x+10=0

step 3: factor left side of equation.

−2(2x+5)(x−1)=0

step 4: set factors equal to 0.

x=-5/2 or x=1

Answer from: Quest

Hope this ; )

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