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Karo-lina-s [1.5K]
3 years ago
14

Kent mixed oil and gas for his lawn mower. He mixed 8 fluid ounces of oil for every 1 gallon of gas. Which of the following stat

ements must be true?
​Select each true statement.
A
For every 2 gallons of gas, he used 16 fluid ounces of oil.
B
For every 3 gallons of gas, he used 38 fluid ounces of oil.
C
For every 5 gallons of gas, he used 40 fluid ounces of oil.
D
For every fluid ounce of oil, he used 18 gallon of gas.
E
For every 40 fluid ounce of oil, he used 10 gallon of gas.
F
For every 10 fluid ounce of oil, he used 114 gallon of gas.
Mathematics
1 answer:
Reil [10]3 years ago
8 0

Answer:

Step-by-step explanation:

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2 years ago
the half-life of chromium-51 is 38 days. If the sample contained 510 grams. How much would remain after 1 year?​
madam [21]

Answer:

About 0.6548 grams will be remaining.  

Step-by-step explanation:

We can write an exponential function to model the situation. The standard exponential function is:

f(t)=a(r)^t

The original sample contained 510 grams. So, a = 510.

Each half-life, the amount decreases by half. So, r = 1/2.

For t, since one half-life occurs every 38 days, we can substitute t/38 for t, where t is the time in days.

Therefore, our function is:

\displaystyle f(t)=510\Big(\frac{1}{2}\Big)^{t/38}

One year has 365 days.

Therefore, the amount remaining after one year will be:

\displaystyle f(365)=510\Big(\frac{1}{2}\Big)^{365/38}\approx0.6548

About 0.6548 grams will be remaining.  

Alternatively, we can use the standard exponential growth/decay function modeled by:

f(t)=Ce^{kt}

The starting sample is 510. So, C = 510.

After one half-life (38 days), the remaining amount will be 255. Therefore:

255=510e^{38k}

Solving for k:

\displaystyle \frac{1}{2}=e^{38k}\Rightarrow k=\frac{1}{38}\ln\Big(\frac{1}{2}\Big)

Thus, our function is:

f(t)=510e^{t\ln(.5)/38}

Then after one year or 365 days, the amount remaining will be about:

f(365)=510e^{365\ln(.5)/38}\approx 0.6548

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