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4 years ago

A paint can has a radius of 4 inches and a height of 15 inches. What is the volume of the paint can?

Mathematics

2 answers:

xxMikexx [17]4 years ago

7 0

For this case we have that by definition, the volume of a cylinder is given by:

$V = \pi * r ^ 2 * h$

Where:

A: It's the radio

h: It's the height

We have according to the data:

$r = 4 \ in\\h = 15 \ in$

Substituting:

$V = \pi * (4) ^ 2 * 15\\V = \pi * 16 * 15\\V = \pi * 240\\V = 753.6 \ in$

Thus, the volume of the paint can is 753.6 cubic inches.

Answer:

$V = 753.6 \ in^3$

fenix001 [56]4 years ago

4 0

The answer is 753.6 inches.

Hope this helps!

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Lana71 [14]

Answer:

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Step-by-step explanation:

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$\huge \boxed{\mathbb{QUESTION} \downarrow}$

What is $\tt\:y+3y=0+24$ ? Explain it.

$\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}$

$\tt \: y + 3 y = 0 + 24$

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3 years ago

On a road map with a scale of 1 cm. : 110 mi., the distance between two cities measures 4.5 cm. What is the actual distance betw

Scorpion4ik [409]

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3 years ago

A consumer products company is formulating a new shampoo and is interested in foam height (in mm). Foam height is approximately

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Answer:

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b) 0.5234

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Step-by-step explanation:

To find the p-value if the sample average is 185, we first compute the z-score associated to this value, we use the formula

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where

$\bar x=mean\; of\;the \;sample$

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N = size of the sample.

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$z=\frac{185-175}{20/\sqrt {10}}=1.5811$

$\boxed {z=1.5811}$

As the sample suggests that the real mean could be greater than the established in the null hypothesis, then we are interested in the area under the normal curve to the right of 1.5811 and this would be your p-value.

We compute the area of the normal curve for values to the right of 1.5811 either with a table or with a computer and find that this area is equal to 0.0569 = 0.057 rounded to 3 decimals.

So the p-value is

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Since the z-score associated to an α value of 0.05 is 1.64 and the z-score of the alternative hypothesis is 1.5811 which is less than 1.64 (z critical), we cannot reject the null, so we are making a Type II error since 175 is not the true mean.

We can compute the probability of such an error following the next steps:

Step 1

Compute $\bar x_{critical}$