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The population of Tinyville is 16. 2 Tinys per acre. Exactly 405 Tinys live in Tinyville. How many acres is Tinyville? Enter onl

crimeas [40]

The area of Tinyville if the population of Tinyville is 405 Tinys while the population density of the Tinyville is 16.2 is 25 Acres.

<h3>What is Ratio?</h3>

A ratio shows us the number of times a number contains another number.

As it is given that the population density of Tinyville is 16.2 Tinys per acre, while the population of Tinyville is 405 Tinys. Therefore, the area of the Tinyville is the ratio of the population of Tinyville to the population density of the Tinyville.

$\text{Area of Tinyville} = \dfrac{\text{Population of Tinyville}}{\text{Population Density of Tinyville}}$

$\text{Area of Tinyville} = \dfrac{405}{16.2} = 25$

Hence, the area of Tinyville if the population of Tinyville is 405 Tinys while the population density of the Tinyville is 16.2 is 25 Acres.

Learn more about Ratio:

brainly.com/question/1504221

6 0

3 years ago

Explain in your own words what a dilation is.

Delicious77 [7]

Answer:

Dilation is the action of becoming or being made wider, larger, or more open.

hope this helps:)

plz can i get brainliest

7 0

3 years ago

A student wanted to construct a 95% confidence interval for the mean age of students in her statistics class. She randomly selec

VMariaS [17]

Answer:

$19.1-3.355\frac{1.5}{\sqrt{9}}=17.42$

$19.1+3.355\frac{1.5}{\sqrt{9}}=20.78$

And the best option would be:

C. [17.42,20.78]

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=19.1$ represent the sample mean

$\mu$ population mean (variable of interest)

s=1.5 represent the sample standard deviation

n=9 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=9-1=8$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,8)".And we see that $t_{\alpha/2}=$

Now we have everything in order to replace into formula (1):

$19.1-3.355\frac{1.5}{\sqrt{9}}=17.42$

$19.1+3.355\frac{1.5}{\sqrt{9}}=20.78$

And the best option would be:

C. [17.42,20.78]

5 0

3 years ago

Multiple the binomials (x+3)^2

maxonik [38]

Answer:

$=x^2+6x+9$

Step-by-step explanation:

$\left(x+3\right)^2\\\mathrm{Apply\:Perfect\:Square\:Formula}:\quad \left(a+b\right)^2=a^2+2ab+b^2\\a=x,\:\:b=3\\=x^2+2x\cdot \:3+3^2\\\mathrm{Simplify}\:x^2+2x\cdot \:3+3^2:\quad x^2+6x+9\\x^2+2x\cdot \:3+3^2\\\mathrm{Multiply\:the\:num\\bers:}\:2\cdot \:3=6\\=x^2+6x+3^2\\3^2=9\\=x^2+6x+9$

8 0

3 years ago

Write a trinomial which is degree 2, where the coefficient of the highest degree term is 7.

Nady [450]

Answer:

7x^2 - 2x - 3.

Step-by-step explanation:

Starts with 7x^2 then a term in x then a constant.

4 0

3 years ago