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Yuliya22 [10]
3 years ago
11

The amount of money in an account with continuously compounded interest is given by the formula A = Pert, where P is the princip

al, r is the annual interest rate, and t is the time in years. Calculate to the nearest tenth of a year how long it takes for an amount of money to double if interest is compounded continuously at 5.2%.
Mathematics
1 answer:
Mama L [17]3 years ago
5 0
Hi there

The formula in this situation is
2=e^0.052t
Solve for t
T=(log(2)÷log(e))÷0.052
T=13.3 years round 13 years

Hope it helps
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ankoles [38]
L: w

Equation

2/3 = 14/ x

Cross multiply

2xX = 14x3

2x = 42

Eliminate

2x/2 = 42/2

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Answer in simplest form
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3 years ago
Emily would like to buy some new
bonufazy [111]

Answer:

$14,277.80

Step-by-step explanation:

The standard formula for compound interest is given as;

A = P(1+r/n)^(nt) .....1

Where;

A = final amount/value

P = initial amount/value (principal)

r = rate yearly

n = number of times compounded yearly.

t = time of investment in years

For this case;

P = $7,400

t = 8 years

n = 4 (quarterly)

r = 9.5% = 0.095

Using equation 1.

A = $7,400(1+0.095/4)^(4×7)

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6 0
2 years ago
Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be
Taya2010 [7]

Answer

Josh's textbook reached the ground first

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

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>

Thus we have for Ben, Eqn. (3) gives:

h(t)=0-16t^2+60t+40=0

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

Since time can only be positive we reject the t_{2} solution and we keep that Ben's book took t=4.3276 seconds to reach the ground.

Similarly solving for Josh we obtain

t_{1}=3.6794sec\\t_{2}=-0.6794sec

Thus again we reject the negative and keep the positive solution, so Josh's book took t=3.6794 seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482

So to answer the original question:

<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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