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3 years ago

The diameter of a round rug is 42 inches. What is the approximate area of the rug?

Mathematics

1 answer:

aksik [14]3 years ago

3 0

Answer:

D, It would be 1,385 Square Inches

Step-by-step explanation:

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Y dont know the answer, can you please help me?

sukhopar [10]

Answer:

it is a

Step-by-step explanation:

3 0

2 years ago

Can someone add these up and tell me what my final grades would be

denis23 [38]

Answer:

yea sure

Step-by-step explanation:

math pretty sure its A

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4 0

2 years ago

Read 2 more answers

Mel has to put the greatest number of

Dmitry_Shevchenko [17]

Mel should use the least common multiple to solve the problem

<u>Solution:</u>

Given, Mel has to put the greatest number of bolts and nuts in each box so each box has the same number of bolts and the same number of nuts.

We have to find that should Mel use the greatest common factor or the least common multiple to solve the problem?

He should use least common multiple.

Let us see an example, suppose 12 bolts and nuts are to be fit in 6 boxes.

Then, if we took H.C.F of 12 and 6, it is 6, which means 6 bolts and nuts in each box, but, after filling 2 boxes with 6 bolts and nuts, there will be nothing left, which is wrong as remaining boxes are empty.

So the remaining method to choose is L.C.M.

Hence, he should use L.C.M method.

6 0

3 years ago

I like turtles so how fast can a turtle cross the street if it was going 2 mils per hour and 2,934,000 hours past?

Studentka2010 [4]

Answer:

5,868,000

Step-by-step explanation:

2,934,000 x 2 = 5,868,000

3 0

3 years ago

Intravenous fluid bags are filled by an automated filling machine. Assume that the fill volumes of the bags are independent, nor

kenny6666 [7]

Answer:

a) 0.0167

b) 0

c) 5.948

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 6.16 ounces

Standard Deviation, σ = 0.08 ounces

We are given that the distribution of fill volumes of bags is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) Standard deviation of 23 bags

$\displaystyle\frac{S.D}{\sqrt{23}} = \frac{0.08}{\sqrt{23}} = 0.0167$

b) P( fill volume of 23 bags is below 5.95 ounces)

P(x < 5.95)

$P( x < 5.96) = P( z < \displaystyle\frac{5.95 - 6.16}{0.0167}) = P(z < -12.57)$

$= 1 - P(z < 12.57)$

Calculation the value from standard normal z table, we have,

$P(x < 5.95) = 1 - 1 = 0$

c) P( fill volume of 23 bags is below 6 ounces) = 0.001

P(x < 6) = 0.001

$P( x < 6) = P( z < \displaystyle\frac{6 - \mu}{0.0167})$

Calculation the value from standard normal z table, we have,

$P( z \leq -3.09) = 0.001$

$\displaystyle\frac{6 - \mu}{0.0167} = -3.09\\\\\mu = 6 + (0.0167\times -3.09) = 5.948$

If the mean will be 5.948 then the probability that the average of 23 bags is below 6.1 ounces is 0.001.

7 0

3 years ago