36.78 divided by 2.4 show all work

A town has a population of 12000 and grows at 3.5% every year. What will be the population after 7 years, to the nearest whole n

Elena L [17]

Answer:

After 7 years the population will be 15268.

Step-by-step explanation:

Given a town has a population of 12000 and grows at 3.5% every year. we have to find the population after 7 years.

As the general growth formula is

$A = P(1 + r)^t$

where A = Amount after growth

P = Original amount

r = yearly growth rate

t = no. of years

here, P=12000, r=0.035, t=7

$A=12000(1+0.035)^7$

= $12000(1.035)^7$

= $15267.35\sim15268$

Hence, after 7 years the population will be 15268.

3 0

3 years ago

a fruit salad is made from pineapples, pears, and peaches mixed in the ratio of 2 to 3 to 5, respectively, by weight. what fract

musickatia [10]

2/10 or 1/5, which is 20%

6 0

4 years ago

Pls answer with formula. will rate 5.0

lara31 [8.8K]

Answer:

a) -19/60

Step-by-step explanation:

add 7/12 to -19/60 and you get 0.26 repeating.

4/15 is 0.26 repeating.

so yea hope this helped :)

6 0

3 years ago

Read 2 more answers

Is this a right triangle?

miskamm [114]

Answer:

yes

Step-by-step explanation:

when 6 and 5 intersect it makes a ninty degree angle which is also known as a right angle and since its a triangle it simplifies to a right triangle. Sorry if i am wrong but im pretty sure im right lol

3 0

3 years ago

Read 2 more answers

It currently takes users a mean of 25 minutes to install the most popular computer program made by RodeTech, a software design c

ArbitrLikvidat [17]

Testing the hypothesis in this problem which is a two-tailed test, we can conclude that there is not sufficient evidence to conclude that the mean time of installation has changed, since the p-value of the test is 0.0364 > 0.01,

At the null hypothesis, we test if the mean is of 25 minutes, that is:

$H_0: \mu = 25$

At the alternative hypothesis, we test if the mean has changed, that is, if it is different than 25 minutes.

$H_1: \mu \neq 25$

We have the standard deviation for the sample, thus, the t-distribution is used. The value of the test statistic is:

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

In which:

X is the sample mean.
$\mu$ is the value tested at the null hypothesis.
s is the standard deviation of the sample.
n is the sample size.

In this problem:

25 is tested at the null hypothesis, thus $\mu = 25$ .
Sample mean of 26.2 minutes, thus $X = 26.2$ .
Sample of 51, thus $n = 51$ .
Variance of 16, thus $s = \sqrt{16} = 4$ .

The value of the test statistic is:

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

$t = \frac{26.2 - 25}{\frac{4}{\sqrt{51}}}$

$t = 2.14$

The p-value of the test is found using a two-tailed test(test if the mean is different of a value), with t = 2.14 and 50 degrees of freedom.

Using a t-distribution calculator, this p-value is of 0.0364.

Since the p-value of the test is 0.0364 > 0.01, there is not sufficient evidence to conclude that the mean time of installation has changed.

A similar problem is given at brainly.com/question/23777908

3 0

3 years ago