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lord [1]
3 years ago
5

A pair of pants regularly costs $68. The pants are on sale for 45% of the original price. How much will the discount be?

Mathematics
1 answer:
Novay_Z [31]3 years ago
6 0

The discounted amount of pair of pants is $37.4

<u>Solution:</u>

Given, A pair of pants regularly costs $68.  

The pants are on sale for 45% of the original price.  

We have to find that how much will the discount be?

Now, <em>discounted amount = original price – sold price </em>

Discounted amount = original price – 45% of original price

Discounted amount = $68 – 45% of $68

\begin{array}{l}{=68-\frac{45}{100} \times 68=68\left(1-\frac{45}{100}\right)=68 \times \frac{100-45}{100}=68 \times \frac{55}{100}} \\\\ {=\frac{3740}{100}=\$ 37.4}\end{array}

Hence, the discounted amount is $37.4

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Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca
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Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

\frac{dQ}{dt} = -rt

Which has the following solution:

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

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that Q(6) = (1-0.03)Q(0) = 0.97Q(0). We use this to find r.

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

0.97Q(0) = Q(0)e^{-6r}

e^{-6r} = 0.97

\ln{e^{-6r}} = \ln{0.97}

-6r = \ln{0.97}

r = -\frac{\ln{0.97}}{6}

r = 0.0051

So

Q(t) = Q(0)e^{-0.0051t}

Determine the half-life of the radioactive substance.

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

Q(t) = Q(0)e^{-0.0051t}

0.5Q(0) = Q(0)e^{-0.0051t}

e^{-0.0051t} = 0.5

\ln{e^{-0.0051t}} = \ln{0.5}

-0.0051t = \ln{0.5}

t = -\frac{\ln{0.5}}{0.0051}

t = 135.9

The half-life of the radioactive substance is 135.9 hours.

6 0
3 years ago
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