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Papessa [141]
3 years ago
7

An amount of $15000 is borrowed for 13 years at 3.25% interest, compounded annually. If the loan is paid in full at the end of t

hat period, how much must be paid back?
Mathematics
1 answer:
kvasek [131]3 years ago
5 0

Answer:

22733.28$(This is a rounded approximate answer)

Step-by-step explanation:

The equation for exponential growth is A = P(1 + r)^n where A is the total money(interest), P is the principal, r is the rate(compound interest rate) and n is the amount of time. If you look closely and read carefully, you will find out that the principle is $15000. The rate is 0.0325(converted into decimal) and the time is 13 years. If you plug in all of this, you should get A = 15000(1.0325)^13, and A will equal approximately 22733.28$.

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Step-by-step explanation:

All you need to do is divide cost by mile since they already factored in the additional $3 for you.

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2 years ago
Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The price of these bulbs is very favorab
Katyanochek1 [597]

Answer:

(a) Customer will not purchase the light bulbs at significance level of 0.05

(b) Customer will purchase the light bulbs at significance level of 0.01 .

Step-by-step explanation:

We are given that Light bulbs of a certain type are advertised as having an average lifetime of 750 hours. A random sample of 50 bulbs was selected,  and the following information obtained:

Average lifetime = 738.44 hours and a standard deviation of lifetimes = 38.2 hours.

Let Null hypothesis, H_0 : \mu = 750 {means that the true average lifetime is same as what is advertised}

Alternate Hypothesis, H_1 : \mu < 750 {means that the true average lifetime is smaller than what is advertised}

Now, the test statistics is given by;

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

where, X bar = sample mean = 738.44 hours

               s  = sample standard deviation = 38.2 hours

               n = sample size = 50

So, test statistics = \frac{738.44-750}{\frac{38.2}{\sqrt{50} } } ~ t_4_9

                            = -2.14

(a) Now, at 5% significance level, t table gives critical value of -1.6768 at 49 degree of freedom. Since our test statistics is less than the critical value of t, so which means our test statistics will lie in the rejection region and we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is smaller than what is advertised and so consumer will not purchase the light bulbs.

(b) Now, at 1% significance level, t table gives critical value of -2.405 at 49 degree of freedom. Since our test statistics is higher than the critical value of t, so which means our test statistics will not lie in the rejection region and we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is same as what it has been advertised and so consumer will purchase the light bulbs.

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Answer:

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Cancel the common factors or r⁹ and r³

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