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Darina [25.2K]
3 years ago
15

Determine the Median, Mode, and mean. Please don't answer if you don't know

Mathematics
2 answers:
Bas_tet [7]3 years ago
8 0
For each separate graph? If so, for the first one the median is 9. But if for all together the median is 11, mode is also 11 and mean is 13
Soloha48 [4]3 years ago
7 0
The mean is 726!!!!!!!!!!!!!!
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Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li
krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years

First, let us calculate the decay constant (k)

75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036

Next, let us calculate the half-life as follows:

\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2}  \approx 669\ years

Therefore the half-life is between 600 and 700 years

5 0
3 years ago
Evaluate 16^1/2 and show work​
LUCKY_DIMON [66]
Here is your answer

{16}^{1/2} = ({4}^{2})^{1/2}\\= {4}^ {2× 1/2}\\= 4

HOPE IT IS USEFUL
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Answer: 10

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Inessa [10]

Answer:

x=15

Step-by-step explanation:

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