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Yuki888 [10]
3 years ago
5

Stuart put $99 into a CD that pays 2.6% interest, compounded monthly. According to the Rule of 72, approximately how long will i

t take for his money to double? Round to the nearest tenth. Show your work.
Mathematics
2 answers:
Temka [501]3 years ago
6 0

Answer:  27 years (approx)

Step-by-step explanation:

According to the Rule of 72,

If the product of interest rate and time is equal to 72, the amount will be doubled.

According to the question,

The annual rate = 2.6%

Let after t years the amount will be doubled.

Thus, 2.6 × t = 72

t = \frac{72}{2.6} = 27.6923076923

Since, the actual time will always less than the time gets by the 72 rule.

Thus, Approx 27 year after the amount will be doubled.


Taya2010 [7]3 years ago
4 0
72/2.6=27.7 years
Hope it helps :-)
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3 years ago
According to one cosmological theory, there were equal amounts of the two uranium isotopes 235U and 238U at the creation of the
FromTheMoon [43]

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6 billion years.

Step-by-step explanation:

According to the decay law, the amount of the radioactive substance that decays is proportional to each instant to the amount of substance present. Let P(t) be the amount of ^{235}U and Q(t) be the amount of ^{238}U after t years.

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                               \frac{dP}{dt} = -k_1P \quad \frac{dQ}{dt} = -k_2Q

where k_1 and k_2 are proportionality constants and the minus signs denotes decay.

Rearranging terms in the equations gives

                             \frac{dP}{P} = -k_1dt \quad \frac{dQ}{Q} = -k_2dt

Now, the variables are separated, P and Q appear only on the left, and t appears only on the right, so that we can integrate both sides.

                         \int \frac{dP}{P} = -k_1 \int dt \quad \int \frac{dQ}{Q} = -k_2\int dt

which yields

                      \ln |P| = -k_1t + c_1 \quad \ln |Q| = -k_2t + c_2,

where c_1 and c_2 are constants of integration.

By taking exponents, we obtain

                     e^{\ln |P|} = e^{-k_1t + c_1}  \quad e^{\ln |Q|} = e^{-k_12t + c_2}

Hence,

                            P  = C_1e^{-k_1t} \quad Q  = C_2e^{-k_2t},

where C_1 := \pm e^{c_1} and C_2 := \pm e^{c_2}.

Since the amounts of the uranium isotopes were the same initially, we obtain the initial condition

                                 P(0) = Q(0) = C

Substituting 0 for P in the general solution gives

                         C = P(0) = C_1 e^0 \implies C= C_1

Similarly, we obtain C = C_2 and

                                P  = Ce^{-k_1t} \quad Q  = Ce^{-k_2t}

The relation between the decay constant k and the half-life is given by

                                            \tau = \frac{\ln 2}{k}

We can use this fact to determine the numeric values of the decay constants k_1 and k_2. Thus,

                     4.51 \times 10^9 = \frac{\ln 2}{k_1} \implies k_1 = \frac{\ln 2}{4.51 \times 10^9}

and

                     7.10 \times 10^8 = \frac{\ln 2}{k_2} \implies k_2 = \frac{\ln 2}{7.10 \times 10^8}

Therefore,

                              P  = Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} \quad Q  = Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}

We have that

                                          \frac{P(t)}{Q(t)} = 137.7

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Solving for t yields t \approx 6 \times 10^9, which means that the age of the  universe is about 6 billion years.

5 0
3 years ago
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