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Karolina [17]
3 years ago
10

You have $44000 in a savings account that pays 2% annual interestand the inflatation rate is 3.24%. How much buying power in dol

lars will you lose in one year because of inflation?
Mathematics
1 answer:
puteri [66]3 years ago
4 0
At the end of the year ...
   the price will be 44,000*1.0324 = 45,425.60
   the bank balance will be 44,000*1.02 = 44,880.00

This is 545.60 fewer dollars than required to pay the price.

You lose $545.60 in current-year buying power.
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There are 52 weeks in a year, so subtract 27 from 52.

52-27=25

There were 25 weeks with no rainy days
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In 14b, is 14 the variable, coefficient, term, or product?
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Step-by-step explanation:

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Describe a mathematical idea or philosophy that came out of the Vitruvian Man.
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provides a procedure to hand-draw a square and a circle of equal area

Step-by-step explanation:

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8 0
3 years ago
The half-life of cesium-137 is 30 years. Suppose we have a 180-mg sample. (a) Find the mass that remains after t years. (b) How
DedPeter [7]

Answer:

a) Q(t) = 180e^{-0.023t}

b) 11.4mg of cesium-137 remains after 120 years.

c) 225.8 years.

Step-by-step explanation:

The following equation is used to calculate the amount of cesium-137:

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

In which Q(t) is the amount after t years, Q(0) is the initial amount, and r is the rate at which the amount decreses.

(a) Find the mass that remains after t years.

The half-life of cesium-137 is 30 years.

This means that Q(30) = 0.5Q(0). We apply this information to the equation to find the value of r.

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

0.5Q(0) = Q(0)e^{-30r}

e^{-30r} = 0.5

Applying ln to both sides of the equality.

\ln{e^{-30r}} = \ln{0.5}

-30r = \ln{0.5}

r = \frac{\ln{0.5}}{-30}

r = 0.023

So

Q(t) = Q(0)e^{-0.023t}

180-mg sample, so Q(0) = 180

Q(t) = 180e^{-0.023t}

(b) How much of the sample remains after 120 years?

This is Q(120).

Q(t) = 180e^{-0.023t}

Q(120) = 180e^{-0.023*120}

Q(120) = 11.4

11.4mg of cesium-137 remains after 120 years.

(c) After how long will only 1 mg remain?

This is t when Q(t) = 1. So

Q(t) = 180e^{-0.023t}

1 = 180e^{-0.023t}

e^{-0.023t} = \frac{1}{180}

e^{-0.023t} = 0.00556

Applying ln to both sides

\ln{e^{-0.023t}} = \ln{0.00556}

-0.023t = \ln{0.00556}

t = \frac{\ln{0.00556}}{-0.023}

t = 225.8

225.8 years.

8 0
3 years ago
Read 2 more answers
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