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aleksandrvk [35]

2 years ago

7

Miles is saving to buy a new car. He currently has $600 in his savings account. He plans on depositing $200 a month until he has

saved the $2000 he needs to buy a car. Rate change?

1 answer:

Marizza181 [45]2 years ago

7 0

Answer:

7 months

Step-by-step explanation:

He plans on depositing $200 every month into the savings account that already has $600 in it.

We can represent the amount of money he will have after a certain number of months (x) as:

A = 600 + 200x

He needs to save $2000. Therefore, to find the months he needs to save, we need to find x when A is $2000:

2000 = 600 + 200x

200x = 2000 - 600

200x = 1400

x = 1400/200 = 7 months

He needs to save for 7 months.

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A sample of a radioactive substance decayed to 97% of its original amount after a year. (Round your answers to two decimal place

Andrej [43]

Answer:

a) The half life of the substance is 22.76 years.

b) 5.34 years for the sample to decay to 85% of its original amount

Step-by-step explanation:

The amount of the radioactive substance after t years is modeled by the following equation:

$P(t) = P(0)(1-r)^{t}$

In which P(0) is the initial amount and r is the decay rate.

A sample of a radioactive substance decayed to 97% of its original amount after a year.

This means that:

$P(1) = 0.97P(0)$

Then

$P(t) = P(0)(1-r)^{t}$

$0.97P(0) = P(0)(1-r)^{0}$

$1 - r = 0.97$

So

$P(t) = P(0)(0.97t)^{t}$

(a) What is the half-life of the substance?

This is t for which P(t) = 0.5P(0). So

$P(t) = P(0)(0.97t)^{t}$

$0.5P(0) = P(0)(0.97t)^{t}$

$(0.97)^{t} = 0.5$

$\log{(0.97)^{t}} = \log{0.5}$

$t\log{0.97} = \log{0.5}$

$t = \frac{\log{0.5}}{\log{0.97}}$

$t = 22.76$

The half life of the substance is 22.76 years.

(b) How long would it take the sample to decay to 85% of its original amount?

This is t for which P(t) = 0.85P(0). So

$P(t) = P(0)(0.97t)^{t}$

$0.85P(0) = P(0)(0.97t)^{t}$

$(0.97)^{t} = 0.85$

$\log{(0.97)^{t}} = \log{0.85}$

$t\log{0.97} = \log{0.85}$

$t = \frac{\log{0.85}}{\log{0.97}}$

$t = 5.34$

5.34 years for the sample to decay to 85% of its original amount

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