If the area of the rectangle is 180sq.ft, find its perimeter

A circular pool if filling with water. Assuming the water level will be 4' deep and the diameter is 20' , what is the volume of

Yuliya22 [10]

35,569.92 L of water is required to fill the pool

Step-by-step explanation:

Step 1: Find the area of the circular pool when radius = 20/2 = 10 ft

⇒ Area = πr² = 3.14 × 10² = 314 ft²

Step 2: Find the volume of the pool

Volume = Area × Depth

= 314 × 4 = 1256 ft³

Step 3: Convert the volume into liters.

Volume of water = 1256 × 28.32 = 35,569.92 L (∵1 ft³ = 28.32 L)

8 0

3 years ago

If $10,000 is invested in an account at 4.8% annual simple interest, how long will it take for the account balance

Lesechka [4]

10000+4.8%=10480
10480+4.8%=10983.04
10983.04+4.8%=11510.22592

so 3 years

7 0

3 years ago

Read 2 more answers

Suppose that you had the following data set. 500 200 250 275 300 Suppose that the value 500 was a typo, and it was suppose to be

hodyreva [135]

Answer:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

Step-by-step explanation:

The subindex B is for the before case and the subindex A is for the after case

Before case (with 500)

For this case we have the following dataset:

500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

And the sample deviation with the following formula:

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

After case (With -500 instead of 500)

For this case we have the following dataset:

-500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

And the sample deviation with the following formula:

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

And as we can see we have a significant change between the two values for the two cases.

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

8 0

3 years ago

What are the roots of the equation? x^2(5x−7)(3x+2)=0

lisov135 [29]

x = 0, x = $\frac{7}{5}$ , x = - $\frac{2}{3}$

since we have a product of factors equal to zero, equate each factor to zero and solve for x

x² = 0 ⇒ x = 0 ( multiplicity 2 )

5x - 7 = 0 ⇒ x = $\frac{7}{5}$

3x + 2 = 0 ⇒ x = - $\frac{2}{3}$

7 0

3 years ago

Read 2 more answers

Find the surface area of the regular pyramids. Round your answer to the nearest tenth.

katrin2010 [14]

Answer:

S.A = 147.35 cm²

Step-by-step explanation:

Surface area of the triangular pyramid = base area + ½(perimeter of base*slant height)

Base area = ½*bh

b = 7 cm

h = 6.1 cm

Base area = ½*7*6.1 = 21.35 cm²

Perimeter of triangular base = 7 + 7 + 7 = 21 cm

slant height = 12 cm

Plug in the values into the equation

S.A = 21.35 + ½(21*12)

S.A = 21.35 + 126

S.A = 147.35 cm²

8 0

2 years ago