Answer:
The land which are used for exhibits are <u>89 acres</u> (approximately).
Step-by-step explanation:
Given:
Village is spread across 240 acres of land. 37% of land is used for the exhibits.
Now, we have to find acres of land which are used for exhibits.
So, here it is given that land used for exhibits is 37%, and we have to calculate the quantity from the percentage given:
According to question:
37% of 240 acres of land.



So, the land used for exhibits are 88.80 acres.
Therefore, the land which are used for exhibits are <u>89 acres </u>(approximately).
A
Don't forget that you can also pull out an x besides the 10. The expression should look like this.
A(x) = 10x(x - 5)
The GCF is 10x
B
The area of the Bedroom = 10x ( x - 5)
If the length = 5x then the width is
A/L = 10x (x - 5)/(5x)= 2*(x - 5)
C
Now they are changing the Length to the GCF. We determined in A that the GCF = 10x
This time A/L = 10x(x - 5)/10x = (x - 5)
So the width = x - 5
Hi there!
Before we begin, let's rewrite your equation. :)

Step 1) Add 14 to both sides.

Step 2) Divide both sides by 5.


Final Answer -

Hope this helps!
Message me if you need anymore help! :D
True because one has a whole number the other is less then a whole number hope this helped ya :)
Answer:

And the standard error is given by:

And replacing we got:
![SE_[p]= \sqrt{\frac{0.06*(1-0.06)}{373}}= 0.0123](https://tex.z-dn.net/?f=%20SE_%5Bp%5D%3D%20%5Csqrt%7B%5Cfrac%7B0.06%2A%281-0.06%29%7D%7B373%7D%7D%3D%200.0123)
And we want to find this probability:

We can calculate the z score for this case and we got:

And using the normal distribution table or excel we got:

Step-by-step explanation:
Previous concepts
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".
The population proportion have the following distribution
Solution to the problem
For this case we can find the mean and standard error for the sample proportion with these formulas:

And the standard error is given by:

And replacing we got:
![SE_[p]= \sqrt{\frac{0.06*(1-0.06)}{373}}= 0.0123](https://tex.z-dn.net/?f=%20SE_%5Bp%5D%3D%20%5Csqrt%7B%5Cfrac%7B0.06%2A%281-0.06%29%7D%7B373%7D%7D%3D%200.0123)
And we want to find this probability:

We can calculate the z score for this case and we got:

And using the normal distribution table or excel we got:
