Find the exact area and circumference of a circle whose diameter is 8 meters. Show your work.

It takes your garden hose 20 seconds to fill your 2-gallon watering can. How long will it take to fill a circular inflatable poo

Naddik [55]

Answer:

It will take ≈ 8.27 hours to fill the circular inflatable pool using the garden hose

Step-by-step explanation:

1. Let's review the information provided to us to answer the question correctly:

Time it takes your garden hose to fill a 2-gallon watering can = 20 seconds

Circular inflatable pool size = 13 feet in diameter and 3 feet deep

2. How long will it take to fill a circular inflatable pool?

A. For answering the question, the first step is to calculate the volume of the inflatable pool, this way:

Diameter = 13 feet ⇒ Radius = 6.5 feet

Height = 3 feet

Volume of the inflatable pool = π * r² * h

Replacing with the real values, we have:

Volume of the inflatable pool = 3.1416 * 6.5² * 3

Volume of the inflatable pool = 398.2 feet³

B. Now we need to calculate how much water in gallons the inflatable pool will hold, this way:

Volume of the inflatable pool in gallons = Volume of the inflatable pool in cubic feet * 7.48

Replacing with the real values, we have:

Volume of the inflatable pool in gallons = 398.2 * 7.48

Volume of the inflatable pool in gallons = 2,978.54

C. Finally, we need to calculate the time it will take to fill the pool using the garden hose, this way:

Time it takes your garden hose to fill a 2-gallon watering can = 20 seconds

Simplifying, we have:

Time it takes your garden hose to fill a 1-gallon watering can = 10 seconds

Time it will take to fill the pool with the garden hose = Volume of the inflatable pool in gallons * 10

Replacing with the real values, we have:

Time it will take to fill the pool with the garden hose = 2,978.54 * 10

Time it will take to fill the pool with the garden hose = 29,785.4 seconds or 8.27 hours

It will take ≈ 8.27 hours to fill the circular inflatable pool using the garden hose

6 0

3 years ago

Read 2 more answers

A line of best fit is drawn for the set of points shown on the graph. Which point is an approximate extrapolation for x = 30 fro

Goshia [24]

Answer with explanation:

Equation of Linear Regression line is

y=1.14 x + 3.805

A line of regression is that line which passes through none of the points, few points or all the points in the two dimensional coordinate plane.It tells value of one variable when another variable is known.

for, x=30

y=1.14 × 30 +3.805

y=34.20 +3.805

y=38.005

y=38 (approx)

So, Ordered pair is , (30,38)

Option B

8 0

3 years ago

Read 2 more answers

point P is located at (-2, 7), and point R is located at (1, 0). find the y value for the point Q that is located 2/3 the distan

natulia [17]

Answer: 7/3

Step-by-step explanation:

R-P= (3,-7). the displacement vector.

P+x(R-P) moves to x times distance between

P+(R-P) = R moves to 100% of the distance.

Q = P+2/3(R-P)

= (-2,7)+2/3(3,-7)

= (-2,7)+(2,-14/3)

= (0,7-14/3)

= (0,7/3)

= (0,2.3333...)

7 0

3 years ago

Complete each congruence statement by naming the corresponding angle or side (odd questions only)

Oksi-84 [34.3K]

Given:

1. $\Delta HJI\cong \Delta IQP$

3. $\Delta JLK\cong \Delta FGH$

To find:

The missing values to complete the congruence statements:

1. $\overline{IH}\cong \_\_\_\_$

3. $\angle KJL\cong \_\_\_\_\_$

Solution:

1.

We have,

$\Delta HJI\cong \Delta IQP$

Here, vertices H, J, I are corresponding to I, Q, P. So,

$\overline{IH}\cong \overline{PI}$

Therefore, the complete statement is $\overline{IH}\cong \overline{PI}$ .

3.

We have,

$\Delta JLK\cong \Delta FGH$

Here, vertices J, L, K are corresponding to F, G, H. So,

$\angle KJL\cong \angle HFG$

Therefore, the complete statement is $\angle KJL\cong \angle HFG$ .

4 0

3 years ago

9. On a game show, there are 16 questions: 8 easy, 5 medium-hard, and 3 hard. If contestants are given questions

just olya [345]

Answer:

$\frac{7}{30}\approx0.23$

Step-by-step explanation:

Given:

Number of questions (N) = 16

Number of easy questions (E) = 8

Number of medium-hard questions (M) = 5

Number of hard questions (H) = 3

Now, the probability of getting the first question as easy question is given as:

$P(E1)=\frac{\textrm{Number of easy questions}}{\textrm{Total questions}}\\\\P(E1)=\frac{E}{N}=\frac{8}{16}=0.5$

Now, probability of getting the second question as easy question is given as:

$P(E2)=\frac{\textrm{Number of easy questions left}}{\textrm{Total questions left}}\\\\P(E2)=\frac{E-1}{N-1}=\frac{7}{15}$

Now, probability that the first two contestants will get easy questions is given by the product of $P(E1)\ and\ P(E2)$ . So,

$P(2\ easy\ questions)=P(E1)\times P(E2)\\\\P(2\ easy\ questions)=\frac{1}{2}\times \frac{7}{15}\\\\P(2\ easy\ questions)=\frac{7}{30}\approx 0.23$

Therefore, the probability that the first two contestants will get easy questions is $\frac{7}{30}\approx0.23$

7 0

3 years ago