3 years ago

To which płace would you round to estimate 6,904-4,111 explain

Mathematics

1 answer:

Rainbow [258]3 years ago

8 0

To estimate you would round the numbers, so it would be 7,000 - 4,000 = 3,000.

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half of the pool would be filled in one hour

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3 years ago

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g100num [7]

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3 years ago

The height of a doorway is 8 yards. What is the height of the doorway in inches?

charle [14.2K]

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3 years ago

If there where 10 apples and my frend took 6 how many do i have

hichkok12 [17]

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3 years ago

At the beginning of an experiment, a scientist has 300 grams of radioactive goo. After 150 minutes, her sample has decayed to 37

monitta

Answer:

Half-life of the goo is 49.5 minutes

$G(t)= 300e^{-0.014t}$

191.7 grams of goo will remain after 32 minutes

Step-by-step explanation:

Let $M_0\,,\,M_f$ denotes initial and final mass.

$M_0=300\,\,grams\,,\,M_f=37.5\,\,grams$

According to exponential decay,

$\ln \left ( \frac{M_f}{M_0} \right )=-kt$

Here, t denotes time and k denotes decay constant.

$\ln \left ( \frac{M_f}{M_0} \right )=-kt\\\ln \left ( \frac{37.5}{300} \right )=-k(150)\\-2.079=-k(150)\\k=\frac{2.079}{150}=0.014$

So, half-life of the goo in minutes is calculated as follows:

$\ln \left ( \frac{50}{100} \right )=-kt\\\ln \left ( \frac{50}{100} \right )=-(0.014)t\\t=\frac{-0.693}{-0.014}=49.5\,\,minutes$

Half-life of the goo is 49.5 minutes

$\ln \left ( \frac{M_f}{M_0} \right )=-kt\Rightarrow M_f=M_0e^{-kt}$

So,

$G(t)= M_f=M_0e^{-kt}$

Put $M_0=300\,\,grams\,,\,k=0.014$

$G(t)= 300e^{-0.014t}$

Put t = 32 minutes

$G(32)= 300e^{-0.014(32)}=300e^{-0.448}=191.7\,\,grams$

7 0

3 years ago