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Hatshy [7]
4 years ago
9

Suppose that 11 inches of wire costs 66 cents. at the same rate, how much (in cents) will 4 inches of wire cost?

Mathematics
1 answer:
Andre45 [30]4 years ago
7 0
24 cents.

11/66= 6
4/x= 6
4•6= 24
(proportions)

hope this is clear :))
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Suppose an unknown radioactive substance produces 8000 counts per minute on a Geiger counter at a certain time, and only 500 cou
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Answer:

The half-life of the radioactive substance is of 3.25 days.

Step-by-step explanation:

The amount of radioactive substance is proportional to the number of counts per minute:

This means that the amount is given by the following differential equation:

\frac{dQ}{dt} = -kQ

In which k is the decay rate.

The solution is:

Q(t) = Q(0)e^{-kt}

In which Q(0) is the initial amount:

8000 counts per minute on a Geiger counter at a certain time

This means that Q(0) = 8000

500 counts per minute 13 days later.

This means that Q(13) = 500. We use this to find k.

Q(t) = Q(0)e^{-kt}

500 = 8000e^{-13k}

e^{-13k} = \frac{500}{8000}

\ln{e^{-13k}} = \ln{\frac{500}{8000}}

-13k = \ln{\frac{500}{8000}}

k = -\frac{\ln{\frac{500}{8000}}}{13}

k = 0.2133

So

Q(t) = Q(0)e^{-0.2133t}

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

Q(t) = Q(0)e^{-0.2133t}

0.5Q(0) = Q(0)e^{-0.2133t}

e^{-0.2133t} = 0.5

\ln{e^{-0.2133t}} = \ln{0.5}

-0.2133t = \ln{0.5}

t = -\frac{\ln{0.5}}{0.2133}

t = 3.25

The half-life of the radioactive substance is of 3.25 days.

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Answer:

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Step-by-step explanation:

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