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3 years ago

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A town has a population of 6000 and grows at 3.5% every year. To the nearest year, how long will it be until the population will

reach 7100?

1 answer:

kotykmax [81]3 years ago

7 0

Answer:

5 years.

Step-by-step explanation:

Every year the population increases by a factor 103.5% or 1.035.

The equation relating population and time in years is:

P = 6000(1.035)^t (This is an example of exponential growth)

So we have:

7100 = 6000(1.035)^t

(1.035)^t = 7100/6000

(1.035)^t = 1.1833

t ln 1.035 = ln 1.1833

t = ln 1.1833/ln 1.035

= 4.89.

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Niki makes the same payment every two months to pay off his $61,600 loan. The loan has an interest rate of 9.84%, compounded eve

Alenkasestr [34]

The question is an annuity question with the present value of the annuity given.
The present value of an annuity is given by PV = P(1 - (1 + r/t)^-nt) / (r/t) where PV = $61,600; r = interest rate = 9.84% = 0.0984; t = number of payments in a year = 6; n = number of years = 11 years and P is the periodic payment.
61600 = P(1 - (1 + 0.0984/6)^-(11 x 6)) / (0.0984 / 6)
61600 = P(1 - (1 + 0.0164)^-66) / 0.0164
61600 x 0.0164 = P(1 - (1.0164)^-66)
1010.24 = P(1 - 0.341769) = 0.658231P
P = 1010.24 / 0.658231 = 1534.78
Thus, Niki pays $1,534.78 every two months for eleven years.
The total payment made by Niki = 11 x 6 x 1,534.78 = $101,295.48
Therefore, interest paid by Niki = $101,295.48 - $61,600 = $39,695.48

8 0

3 years ago

Read 2 more answers

2. The time between engine failures for a 2-1/2-ton truck used by the military is

OLEGan [10]

Answer:

A truck "<em>will be able to travel a total distance of over 5000 miles without an engine failure</em>" with a probability of 0.89435 or about 89.435%.

For a sample of 12 trucks, its average time-between-failures of 5000 miles or more is 0.9999925 or practically 1.

Step-by-step explanation:

We have here a <em>random variable</em> <em>normally distributed</em> (the time between engine failures). According to this, most values are around the mean of the distribution and less are far from it considering both extremes of the distribution.

The <em>normal distribution</em> is defined by two parameters: the population mean and the population standard deviation, and we have each of them:

$\\ \mu = 6000$ miles.

$\\ \sigma = 800$ miles.

To find the probabilities asked in the question, we need to follow the next concepts and steps:

We will use the concept of the <em>standard normal distribution</em>, which has a mean = 0, and a standard deviation = 1. Why? With this distribution, we can easily find the probabilities of any normally distributed data, after obtaining the corresponding <em>z-score</em>.
A z-score is a kind of <em>standardized value</em> which tells us the <em>distance of a raw score from the mean in standard deviation units</em>. The formula for it is: $\\ z = \frac{x - \mu}{\sigma}$ . Where <em>x</em> is the value for the raw score (in this case x = 5000 miles).
The values for probabilities for the standard normal distribution are tabulated in the <em>standard normal table</em> (available in Statistics books and on the Internet). We will use the <em>cumulative standard normal table</em> (see below).

With this information, we can solve the first part of the question.

The chance that a truck will be able to travel a total distance of over 5000 miles without an engine failure

We can "translate" the former mathematically as:

$\\ P(x>5000)$ miles.

The z-score for x = 5000 miles is:

$\\ z = \frac{5000 - 6000}{800}$

$\\ z = \frac{-1000}{800}$

$\\ z = -1.25$

This value of z is negative, and it tells us that the raw score is 1.25 standard deviations <em>below</em> the population mean. Most standard normal tables are made using positive values for z. However, since the normal distribution is symmetrical, we can use the following formula to overcome this:

$\\ P(z$

So

$\\ P(z$

Consulting a standard normal table available on the Internet, we have

$\\ P(z$

Then

$\\ P(z1.25)$

$\\ P(z1.25)$

However, this value is for P(z<-1.25), and we need to find the probability P(z>-1.25) = P(x>5000) (Remember that we standardized x to z, but the probabilities are the same).

In this way, we have

$\\ P(z>-1.25) = 1 - P(z$

That is, the complement of P(z<-1.25) is P(z>-1.25) = P(x>5000). Thus:

$\\ P(z>-1.25) = 1 - 0.10565$

$\\ P(z>-1.25) = 0.89435$

In words, a truck "<em>will be able to travel a total distance of over 5000 miles without an engine failure</em>" with a probability of 0.89435 or about 89.435%.

We can see the former probability in the graph below.

The chance that a fleet of a dozen trucks will have an average time-between-failures of 5000 miles or more

We are asked here for a sample of <em>12 trucks</em>, and this is a problem of <em>the sampling distribution of the means</em>.

In this case, we have samples from a <em>normally distributed data</em>, then, the sample means are also normally distributed. Mathematically:

$\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

In words, the samples means are normally distributed with the same mean of the population mean $\\ \mu$ , but with a standard deviation $\\ \frac{\sigma}{\sqrt{n}}$ .

We have also a standardized variable that follows a standard normal distribution (mean = 0, standard deviation = 1), and we use it to find the probability in question. That is

$\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ z \sim N(0, 1)$

Then

The "average time-between-failures of 5000" is $\\ \overline{x} = 5000$ . In other words, this is the mean of the sample of the 12 trucks.

Thus

$\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ z = \frac{5000 - 6000}{\frac{800}{\sqrt{12}}}$

$\\ z = \frac{-1000}{\frac{800}{\sqrt{12}}}$

$\\ z = \frac{-1000}{230.940148}$

$\\ z = -4.330126$

This value is so low for z, that it tells us that P(z>-4.33) is almost 1, in other words it is almost certain that for a sample of 12 trucks, its average time-between-failures of 5000 miles or more is almost 1.

$\\ P(z$

$\\ P(z$

$\\ P(z$

The complement of P(z<-4.33) is:

$\\ P(z>-4.33) = 1 - P(z$ or practically 1.

In conclusion, for a sample of 12 trucks, its average time-between-failures of 5000 miles or more is 0.9999925 or practically 1.

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3 years ago

Bailey likes the shape of her friend's pool, but she wants one that is half the volume. According to Cavalieri’s Principle, what

Cerrena [4.2K]

Answer: I need a picture off the pool and measurements.

Step-by-step explanation: I need this in order to figure out the problem and give you a helpful answer.

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3 years ago

Read 2 more answers

A study was conducted to compare the proportion of drivers in Boston and New York who wore seat belts while driving. Data were c

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(A.)test statistic for this test using Ha: p8くPN is Z = -2.164 (B.) p-value (3 decimal places) is 0.015. (C.) Based on this p-value, which of the following would you expect for a 95% is negative. (D.) the correct conclusion is The proportion of Boston drivers wearing seat was significantly lower than the proportion of new York drivers wearing seat belts.

A.)

According to given data find the test statistic for this test using Ha:

p1 = 0.581

p2 = 0.832

standard error = SE = 0.116

<h3>Test statistics :-</h3>

Z = $\frac{p1 - p2}{SE}$

Z = $\frac{0.581 - 0.832}{0.116}$ = -2.164

Z = -2.164

B.)

According to given Determine the p-value (3 decimal places).

<h3>p - value</h3>

it is left tailed test.

p - value for left tailed test is,

p - value = P(Z < test statistics )

p - value = P(Z < -2.164)

P - value = 0.015

C.)

Based on this p-value, which of the following would you expect for a 95%

confidence level = 95% = 0.95

significance level = α = 0.05

P -value is less than 0.05

so, reject null hypothesis at α = 0.05

That is $P_{B}$ < $P_{NY}$

so, based on P - value , All values in the 95% confidence interval

PB - PNY is negative.

D.)

According to given data What is the correct conclusion?

Reject null hypothesis at α = 0.05

so, $P_{B}$ < $P_{NY}$

Hence, The proportion of Boston drivers wearing seat was significantly lower than the proportion of new York drivers wearing seat belts.

know more about test statics visit this link

brainly.com/question/23332597

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1 year ago

If you divide 6! by 4! you are evaluating P(6, 4)

Aleks [24]

When you're evaluating P(n, k), you are essentially saying there are n objects and you will need to arrange them in k spaces.

This is essentially the same as: $^{6}P_4$ , which follows the formula:

$P(6, 4) = ^{6}P_4 = \frac{6!}{(6 - 4)!} = \frac{6!}{2!}$
If you're dividing by 4!, you are evaluating P(6, 2)

8 0

3 years ago