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3 years ago

7

a large bag contains 12/15 pound of granola. how many 1/3 pund bags van be filled with his amount of granola? how much granola i

s left over?

1 answer:

sertanlavr [38]3 years ago

8 0

Answer:

2 bags filled, left over is 2/15 pounds of granola

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Divide <br> 8 2/5 ÷ (-2 1/5)

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8 2/5 ÷(-2 1/5) = -3.81818181818

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After reading 80% of her e-mails in her inbox, Danette still has M unread emails. Which of the following expressions could repre

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m/(1 - .8)

Step-by-step explanation:

She read 80% of her emails, which is .8 of the total. So her unread emails would be 100% - 80% = 1 - .8

that means me can be written as:

m = (1 - .8)t

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Equations<br> Solve T = C(9+ AB) for B

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Answer:

Simplifying

T = C(9 + AB) * forB

Reorder the terms for easier multiplication:

T = C * forB(9 + AB)

Multiply C * forB

T = forBC(9 + AB)

T = (9 * forBC + AB * forBC)

Reorder the terms:

T = (forAB2C + 9forBC)

T = (forAB2C + 9forBC)

Solving

T = forAB2C + 9forBC

Solving for variable 'T'.

Move all terms containing T to the left, all other terms to the right.

Simplifying

T = forAB2C + 9forBC

Step-by-step explanation:

Simplifying

T = C(9 + AB) * forB

Reorder the terms for easier multiplication:

T = C * forB(9 + AB)

Multiply C * forB

T = forBC(9 + AB)

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2 years ago

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3 years ago

Suppose 52% of the population has a college degree. If a random sample of size 563563 is selected, what is the probability that

amm1812

Answer:

The value is $P(| \^ p - p| < 0.05 ) = 0.9822$

Step-by-step explanation:

From the question we are told that

The population proportion is $p = 0.52$

The sample size is n = 563

Generally the population mean of the sampling distribution is mathematically represented as

$\mu_{x} = p = 0.52$

Generally the standard deviation of the sampling distribution is mathematically evaluated as

$\sigma = \sqrt{\frac{ p(1- p)}{n} }$

=> $\sigma = \sqrt{\frac{ 0.52 (1- 0.52 )}{563} }$

=> $\sigma = 0.02106$

Generally the probability that the proportion of persons with a college degree will differ from the population proportion by less than 5% is mathematically represented as

$P(| \^ p - p| < 0.05 ) = P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 ))$

Here $\^ p$ is the sample proportion of persons with a college degree.

So

$P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P(\frac{[[0.05 -0.52]]- 0.52}{0.02106} < \frac{[\^p - p] - p}{\sigma } < \frac{[[0.05 -0.52]] + 0.52}{0.02106} )$

Here

$\frac{[\^p - p] - p}{\sigma } = Z (The\ standardized \ value \ of\ (\^ p - p))$

=> $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P[\frac{-0.47 - 0.52}{0.02106 } < Z < \frac{-0.47 + 0.52}{0.02106 }]$

=> $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P[ -2.37 < Z < 2.37 ]$

=> $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P(Z < 2.37 ) - P(Z < -2.37 )$

From the z-table the probability of (Z < 2.37 ) and (Z < -2.37 ) is

$P(Z < 2.37 ) = 0.9911$

and

$P(Z < - 2.37 ) = 0.0089$

So

=> $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) =0.9911-0.0089$

=> $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = 0.9822$

=> $P(| \^ p - p| < 0.05 ) = 0.9822$

3 0

3 years ago