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Nimfa-mama [501]
3 years ago
12

You have earned $200 doing chores how long will it take to triple your money if you keep it in an account earning 4.25 compound

continuesly
Mathematics
2 answers:
Alekssandra [29.7K]3 years ago
6 0

Answer:

25.85 years

Step-by-step explanation:

Use the formula for continuous compounding amount:

A = Pe^(r*t), where r is the interest rate as a  decimal fraction, P is the initial amount, and t is the time in years.

Here we have:

A = $200e^(0.0425*t) = 3($200) (this is triple the original amount)

Solve this for t.  Divide both sides by $200 and then use natural logs:

1e^(0.0425*t) = 3

Then 0.0425*t = ln 3 = 1.0986

Dividing both sides by 0.0425 will isolate t:

       1.0986

t = --------------- = 25.85 years

       0.0425

Darina [25.2K]3 years ago
4 0

Answer:

c is the right one

Step-by-step explanation:

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No, at alpha equals 0.10​, we do not have enough evidence to support the​ county's claim.

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We are given that a county is considering raising the speed limit on a road because they claim that the mean speed of vehicles is greater than 30 miles per hour.

A random sample of 15 vehicles has a mean speed of 31 miles per hour and a standard deviation of 4.7 miles per hour.

<em><u>Let </u></em>\mu<em><u> = true mean speed of the vehicles.</u></em>

SO, <u>Null Hypothesis</u>, H_0 : \mu \leq  30 miles per hour   {means that the mean speed of vehicles is lesser than or equal to 30 miles per hour}

<u>Alternate Hypothesis,</u> H_A : \mu > 30 miles per hour   {means that the mean speed of vehicles is greater than 30 miles per hour}

The test statistics that will be used here is <u>One-sample t test statistics</u> as we don't know about the population standard deviation;

                        T.S.  = \frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean speed of 15 vehicles = 31 mph

             s = sample standard deviation = 4.7 mph

             n = sample of vehicles = 15

So, <em><u>test statistics</u></em>  =   \frac{31-30}{\frac{4.7}{\sqrt{15} } }  ~ t_1_4

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<u><em>Hence, the value of test statistics is 0.824.</em></u>

<em />

<em>Now at 0.10 significance level, the t table gives critical value of 1.345 at 14 degree of freedom for right-tailed test. Since our test statistics is less than the critical value of t as 0.824 < 1.345, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.</em>

<em />

Therefore, we conclude that the mean speed of vehicles is lesser than or equal to 30 miles per hour which means that the county's claim is not supported.

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