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WINSTONCH [101]
3 years ago
11

On April 1, 1986, Casey deposited $1150 into a savings account paying 9.6% interest, compounded quarterly. If he hasn't made any

additional deposits or withdrawals since then, and if the interest rate has stayed the same, in what year did his balance hit $2300, according to the rule of 72?
Mathematics
2 answers:
dalvyx [7]3 years ago
7 0
According to the 72 rule
72/rate=time
72÷9.6=7.5 years

Another way to solve by using the main equation
2300=1150(1+0.096/4)^4t
Solve for t
t=((log(2,300÷1,150)÷log(1+(0.096÷4))÷4))=7.31years

Hope it helps :-)
Harman [31]3 years ago
7 0

Answer: 7.5 years.

Step-by-step explanation:

Here the given amount = $1150

Amount after getting compound interest in this amount = $ 2300

Since, $2300 is double to the amount $1150

According to the rule of 72, an amount is doubled if the product of annual rate and period (in years) is equal to 72. Also, with the help of this rule we get the approximate value of rate or time.

Since, Here the annual rate of interest = 9.6%

Let the given amount $1150 is doubled ($2300)  in t years.

The, 9.6 × t = 72

⇒ t = 72/9.6 = 7.5

Thus, given amount $1150 is doubled in 7.5 years(approx).

Verification : Finding the number of year with help of general formula of compound interest.

2300=1150(1+\frac{9.6}{100})^{t}

2=(1+\frac{9.6}{100})^{t}

2=(1+\frac{9.6}{100})^{t}

2^{1/t}=(1+\frac{9.6}{100})

⇒ t = 7.562 years.

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Sarah is driving. Her distance in km from Tempe after t hours of driving is given by: x = D(t) = 13 + 57t
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Answer:

The solutions for your four question problem are:

a) t = D^-1(x) =  (1/57)*x -(13/57)

b) t = D^-1(x) =  1 h

c) D^-1(x) represents time.

d) The number of hours of driving needed for Sarah to be x km from Tempe. (option (c))

Step-by-step explanation:

a) Determine a formula in terms of x for: t = D^-1(x)

The distance in km from Tempe after t hours of driving is given by

x = D(t) = 13 + 57t.

We just need to find the value  t function of x

x = 13 + 57*t

x -13 =  57*t

57*t  = x -13

t = (1/57)*x -(13/57)

We can see the plots of both equation in the picture below.

b) Compute D^-1(70)

Once we find the expression for D^-1(x)

We substitute for x = 70 km

t = D^-1(x) =  (1/57)*x -(13/57)

t = D^-1(x) =  (1/57)*(70) -(13/57)

t = D^-1(x) =  (70/57) -(13/57)

t = D^-1(x) =  (1.228) -(0.228)

t = D^-1(x) = 1 h

c) In the expression D^-1(x) :  what quantity (distance or time) does the x represent?  what quantity (distance or time) does the entire D^-1(x) represent?

x represents Distance in both equations (D(t), and D^-1(x))

t represents Time in both equations (D(t), and D^-1(x))

Since t = D^-1(x),

D^-1(x) represents time.

d) Which of the following statements best describes D^-1(x)?

The number of hours of driving needed for Sarah to be x km from Tempe.

Since, t = D^-1(x), and t represents the amount of time elapsed since Sarah, parted from Tempe, the correct answer is option (c)

The expression for D^-1(x) can be found in the previous answers

t = D^-1(x) =  (1/57)*x -(13/57)

The input is x (distance) and the output is t (time)

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