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g100num [7]
3 years ago
15

The half-life of radioactive strontium-90 is approximately 28 years. In 1960, radioactive strontium-90 was released into the atm

osphere during testing of nuclear weapons, and was absorbed into people's bones. How many years does it take until only 15 percent of the original amount absorbed remains?
Mathematics
1 answer:
pentagon [3]3 years ago
6 0

Answer: It will take about 77 years (76.8 years).

Step-by-step explanation: Time is given by the equation: N=No.e^kt. You know that the concentration has to be 15% from the original so N=15 and No=100. To determine k, you have to look at the half-life. k=ln(0.5)/half-life time. For 28 years, k=-0.0247. Replacing it back into the N=No.e^kt equation, the answer will be 76.8 years.

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pashok25 [27]

Answer:

x = 0

y = 4

Step-by-step explanation:

Given

12x - 5y = -20

y = x + 4

Substitute x + 4 for y in the first equation

12x - 5(x + 4) = -20

Expand the bracket

12x -5 X x - 5 x 4 = -20

12x - 5x - 20 = -20

7x - 20 = -20

Add 20 to both sides to eliminate -20 on the left side

7x - 20 + 20 = -20 + 20

7x = 0

Divide both sides by 7 to isolate x

7x/7 = 0/7

x = 0

Substitute 0 for x in either equation to get y

Using equation 2, we have

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3 0
3 years ago
The spinner is divided into a equal sections which two events have the same probability
myrzilka [38]

Answer:

its A

Step-by-step explanation:

7 0
3 years ago
How do you do this ?
salantis [7]
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5 0
4 years ago
How many years (to two decimal places) will it take $15000 to grow to $17500 if it is invested at 8% compounded semi- annually?
VLD [36.1K]

Answer:

1.97 years

Step-by-step explanation:

First, convert R as a percent to r as a decimal

r = R/100

r = 8/100

r = 0.08 per year,

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t = ln(A/P) / n[ln(1 + r/n)]

t = ln(17,500.00/15,000.00) / ( 2 × [ln(1 + 0.08/2)] )

t = ln(17,500.00/15,000.00) / ( 2 × [ln(1 + 0.04)] )

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:D

3 0
3 years ago
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dalvyx [7]

Answer:

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Step-by-step explanation:

7 0
2 years ago
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