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marta [7]
3 years ago
10

If a sum invested gains 10% each year how long will it be before it has doubled its value?

Mathematics
1 answer:
Anika [276]3 years ago
4 0

Using an exponential equation, it is found that it will be 7.3 years before the value is doubled.

The equation for an increasing <u>exponential function</u> is given by:

A(t) = A(0)(1 + r)^t

In which:

  • A(0) is the initial amount.
  • r is the growth rate, as a decimal.

In this problem:

  • Gains 10% each year, thus r = 0.1.

The doubling time is t for which A(t) = 2A(0), then:

A(t) = A(0)(1.1)^t

2A(0) = A(0)(1.1)^t

(1.1)^t = 2

\log{(1.1)^t} = \log{2}

t\log{1.1} = \log{2}

t = \frac{\log{2}}{\log{1.1}}

t = 7.3

It will be 7.3 years before the value is doubled.

A similar problem is given at brainly.com/question/23008760

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Answer:

3.5 is your answer.

Step-by-step explanation:

(7.7x10^-2) divided by (2.2x10^-2

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Hope this helps.

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Which is the completely factored form of 12x^3-60x^2+4x-20​
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(12x²+4)(x-5)

Step-by-step explanation:

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s= -1

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7= -10s-3

+3        +3

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The selling price of a refrigerator is $584. If the markup is 25% of the dealers cost, what is the dealer’s cost of the refriger
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3 years ago
A phone manufacturer wants to compete in the touch screen phone market. He understands that the lead product has a battery life
alisha [4.7K]

Answer:

a)

The null hypothesis is H_0: \mu \leq 10

The alternative hypothesis is H_1: \mu > 10

b-1) The value of the test statistic is t = 1.86.

b-2) The p-value is of 0.0348.

Step-by-step explanation:

Question a:

Test if the battery life is more than twice of 5 hours:

Twice of 5 hours = 5*2 = 10 hours.

At the null hypothesis, we test if the battery life is of 10 hours or less, than is:

H_0: \mu \leq 10

At the alternative hypothesis, we test if the battery life is of more than 10 hours, that is:

H_1: \mu > 10

b-1. Calculate the value of the test statistic.

The test statistic is:

We have the standard deviation for the sample, so the t-distribution is used to solve this question

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

10 is tested at the null hypothesis:

This means that \mu = 10

In order to test the claim, a researcher samples 45 units of the new phone and finds that the sample battery life averages 10.5 hours with a sample standard deviation of 1.8 hours.

This means that n = 45, X = 10.5, s = 1.8

Then

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{10.5 - 10}{\frac{1.8}{\sqrt{45}}}

t = 1.86

The value of the test statistic is t = 1.86.

b-2. Find the p-value.

Testing if the mean is more than a value, so a right-tailed test.

Sample of 45, so 45 - 1 = 44 degrees of freedom.

Test statistic t = 1.86.

Using a t-distribution calculator, the p-value is of 0.0348.

5 0
2 years ago
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