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3 years ago

PLZ I need help

Mathematics

2 answers:

stealth61 [152]3 years ago

8 0

Answer: it will take 4.8 years

Step-by-step explanation:

Initial amount invested is $5000. This means that the principal is

P = 5000

It was compounded annually. This means that it was compounded once in a year. So

n = 1

The rate at which the principal was compounded is 15.6%. So

r = 15.6/100 = 0.156

The formula for compound interest is

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

A = total amount in the account at the end of t years. The amount after t years would be 2 × 5000 = $10000. It means that A = $10000

Therefore

10000 = 5000 (1+0.156/1)^1×t

10000/5000 = (1.156)^t

2 = (1.156)^t

Take log of both sides

log 2 = log (1.156)^t

0.301 = tlog1.156 = 0.063t

t = 0.301/0.063 = 4.8

enot [183]3 years ago

4 0

Answer:

5 years to the nearest year.

Step-by-step explanation:

The formula for Compound Interest with an annual interest payment is

A = P(1 + r/100)^t where A = Amount after t years, P = Initial amount, r = percentage rate.

We have the equation:

10,000 = 5,000(1 + 15.6/ 100)^t

10000 = 5000 * 1.156^t

1.156^t = 2

t ln 1.156 = ln 2

t = ln2 / ln 1.156

t = 4.78

Answer is 5 years.

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Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li

krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

First, let us calculate the decay constant (k)

$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$