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3 years ago

8

Rebecca deposited money into his retirement account that is compounded annually at an interest rate of 7%. Rebecca thought the e

quivalent quarterly interest rate would be 2%. Is Rebecca correct ? If she is, explain why. If she is not correct state what the equivalent quarterly interest rate is and show how you got your answer.

1 answer:

frutty [35]3 years ago

5 0

Answer:

Rebecca is incorrect. The equivalent quarterly interest rate is 1.75%.

Explanation:

The formula for compound interest is i = r/n.

r represents the interest rate in decimal form.

n represents the compounding periods in a year.

Calculate the interest rate for 7% compounded quarterly:

Since interest compounded quarterly is four times a year, n = 4.

7% is converted to decimal form by dividing by 100.

7 / 100 = 0.07

r = 0.07

Substitute these into the formula:

i = r/n

= 0.07/4

= 0.0175

0.0175 is in decimal form. To convert it to a percentage, multiply it by 100.

0.0175 X 100 = 1.75%

1.75% ≠ 2%

Therefore, Rebecca is incorrect. The equivalent quarterly interest rate is 1.75%.

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The answer to your problem is: $x=1/13$

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3 years ago

Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

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Answer:

$A=1500-1450e^{-\dfrac{t}{250}}$

Step-by-step explanation:

The large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved.

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Brine solution with concentration of 2 lb/gal is pumped into the tank at a rate of 3 gal/min

$R_{in}$ =(concentration of salt in inflow)(input rate of brine)

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When the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

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Now, the rate of change of the amount of salt in the tank

$\dfrac{dA}{dt}=R_{in}-R_{out}$

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We solve the resulting differential equation by separation of variables.

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Taking the integral of both sides

$\int(Ae^{\frac{t}{250}})'=\int 6e^{\frac{t}{250}} dt\\Ae^{\frac{t}{250}}=6*250e^{\frac{t}{250}}+C, $(C a constant of integration)\\Ae^{\frac{t}{250}}=1500e^{\frac{t}{250}}+C\\$Divide all through by e^{\frac{t}{250}}\\A(t)=1500+Ce^{-\frac{t}{250}}$

Recall that when t=0, A(t)=50 (our initial condition)

$50=1500+Ce^{-\frac{0}{250}}50=1500+Ce^{0}\\C=-1450\\$Therefore the amount of salt in the tank at any time t is:\\A=1500-1450e^{-\dfrac{t}{250}}$

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