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Lena [83]
2 years ago
6

Jim and bill each invest $15,000 into savings accounts that earn 3.5% interest. Jims account earns simple interests and bills ac

count earns compound interest. After 25 years, who will earn more interest and how much more will he earn.
Mathematics
1 answer:
Svet_ta [14]2 years ago
4 0

Answer:

Bill will earn more interest

He will earn $ 20,448.67 from his investment

Step-by-step explanation:

Firstly let us calculate Jim's earnings based on simple interest

A = P(1 + rt)

Calculation:

First, converting R percent to r

a decimal

r = R/100 = 3.5%/100 = 0.035 per year.

Solving our equation:

A = 15000(1 + (0.035 × 25)) = 28125

A = $28,125.00

The total amount accrued, principal plus interest, from simple interest on a principal of $15,000.00 at a rate of 3.5% per year for 25 years is $28,125.00 for Jim

Now let's us calculate bill's investment based on compound interest

Equation

A = P(1 + r)^t

A=15000(1+0.035)^25

A=15000(1.035)^25

A=15000*2.36324498427

A = $ 35,448.67

We see that Bill will earn

$ 20,448.67 from his investment

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Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

h(t)=-16t^2 +v_{o}t+s       Eqn(1).

Where:

h(t) : is the height range as a function of time

v_{o}   : is the initial velocity

s     : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

h(t)=-16t^2 + v_{o}t + 40        Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

<u>Ben:</u>

We know that v_{o}=60ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+60t+40        Eqn.(3)

<u>Josh:</u>

We know that v_{o}=48ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+48t+40       Eqn. (4).

<em><u>Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at </u></em>h(t)=0<em><u>. Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one. </u></em>

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Using the quadratic expression to find the roots of the quadratic we have:

t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec

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<em>Whose textbook reaches the ground first and by how many seconds?</em>

  • Josh's textbook reached the ground first
  • Josh's textbook reached the ground first by a difference of t=0.6482

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