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3 years ago

5

The groundkeep filled the first sandbox with sand to a height of 6 inches what is the volume in cubic feet of the sand that was

used to kill the first sandbox to a night of 6inches

1 answer:

vodka [1.7K]3 years ago

5 0

Answer:

https://nrckids.org/CFOC/Infant_Toddlers

Step-by-step explanation:

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46

Step-by-step explanation:

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3 years ago

Calculate the length of the circumference of a circle with diameter of 10.4

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What mechanical energy do you transfer to an object when you lift it

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3 years ago

Lamar is writing a coordinate proof to show that a segment from the midpoint of the hypotenuse of a right triangle to the opposi

Zanzabum

1. N is a midpoint of the segment KL, then N has coordinates

$\left(\dfrac{x_K+x_L}{2},\dfrac{y_K+y_L}{2} \right) =\left(\dfrac{0+2a}{2},\dfrac{2b+0}{2} \right) =(a,b).$

2. To find the area of △KNM, the length of the base MK is 2b, and the length of the height is a. So an expression for the area of △KNM is

$A_{KMN}=\dfrac{1}{2}\cdot \text{base}\cdot \text{height}=\dfrac{1}{2}\cdot 2b\cdot a=ab.$

3. To find the area of △MNL, the length of the base ML is 2a and the length of the height is b. So an expression for the area of △MNL is

$A_{MNL}=\dfrac{1}{2}\cdot \text{base}\cdot \text{height}=\dfrac{1}{2}\cdot 2a\cdot b=ab.$

4. Comparing the expressions for the areas you have that the area $A_{KMN}$ is equal to the area $A_{MNL}$ . This means that the segment from the midpoint of the hypotenuse of a right triangle to the opposite vertex forms two triangles with equal areas.

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3 years ago

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Assume that the population proportion is 0.56. Compute the standard error of the proportion, σp, for sample sizes of 100, 200, 5

Aliun [14]

Answer:

Standard errors are 0.049, 0.035, 0.022, and 0.016.

Step-by-step explanation:

The given value of population proportion (P) = 0.56

Given sample sizes (n ) 100, 200, 500, and 1000.

Now standard error is required to calculate.

Use the below formula to find standard error.

When sample size is n = 100

$\sqrt{\frac{P(1-P)}{n}} = \sqrt{\frac{0.56(1-0.56)}{100}} =0.049$

When sample size is n = 200

$\sqrt{\frac{P(1-P)}{n}} = \sqrt{\frac{0.56(1-0.56)}{200}} = 0.035$

When sample size is n = 500

$\sqrt{\frac{P(1-P)}{n}} = \sqrt{\frac{0.56(1-0.56)}{500}} =0.022$

When sample size is n = 1000

$\sqrt{\frac{P(1-P)}{n}} = \sqrt{\frac{0.56(1-0.56)}{1000}} = 0.016$

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3 years ago