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natima [27]
3 years ago
13

A school purchases boxes of candy bars each box contains 50 candy bars each box cost $20. how much does the school have to charg

e for each candy bar to make a profit of $10 per box
Mathematics
2 answers:
Butoxors [25]3 years ago
6 0
20/50=.4  .6x50=30 so ur answer is $0.60
Murrr4er [49]3 years ago
3 0
They would need to charge at least 60 cents per bar

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I got stuck on this question.<br> Use long division to solve
Phantasy [73]

Answer:

  5x² +19x +76 +310/(x-4)

Step-by-step explanation:

The process is straightforward. Find the quotient term, multiply it by the divisor and subtract from the dividend to get the new dividend. Repeat until the dividend is a constant (lower-degree than the divisor).

The tricky part with this one is realizing that there is no x-term in the original dividend, so that term needs to be added with a 0 coefficient. The rather large remainder is also unexpected, but that's the way this problem unfolds.

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Unlike numerical long division, polynomial long division is simplified by the fact that the quotient term is the ratio of the highest-degree terms of the dividend and divisor. Here, the first quotient term is (5x^3)/(x) = 5x^2.

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3 years ago
Write a word problem that the bar model in problem 2 could repereseny
Anna35 [415]
We can’t see the bar model
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2 years ago
What is the volume of a 14 in diameter pipe that’s 2156 long?
storchak [24]

Answer:


Step-by-step explanation:

given that diameter of the pipe = 14 inches

radius = 7 inches

Length =2156 inches

Pipe is in the form of a cylinder

Volume of cylinder = pi r^2 h

where r = radius and h = l = 2156

We have to substitute the values of r and h.

Let us take pi as 22/7

= 22/7 (7)(7)(2156)

Cancel out 7 and multiply the remaining 22 ,7 and 2156

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Evaluate using distributive property <br> 16 x ( -3/8)+16 x 1/4
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8 0
3 years ago
The average student loan debt for college graduates is $25,150.
Aleks04 [339]

Using the normal distribution, it is found that:

a) X \approx N(25150, 12050)

b) There is a 0.5859 = 58.59% probability that the college graduate has between $14,200 and $33,950 in student loan debt.

c) Low: $23,519.65, High: $26,580.35.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

Z = \frac{X - \mu}{\sigma}

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

\mu = 25150, \sigma = 12050.

Hence the distribution of X can be described as follows:

X \approx N(25150, 12050)

The probability that the college graduate has between $14,200 and $33,950 in student loan debt is the <u>p-value of Z when X = 33950 subtracted by the p-value of Z when X = 14200</u>, hence:

X = 33950:

Z = \frac{X - \mu}{\sigma}

Z = \frac{33950 - 25150}{12050}

Z = 0.73

Z = 0.73 has a p-value of 0.7673.

X = 14200:

Z = \frac{X - \mu}{\sigma}

Z = \frac{14200 - 25150}{12050}

Z = -0.91

Z = -0.91 has a p-value of 0.1814.

0.7673 - 0.1814 = 0.5859.

There is a 0.5859 = 58.59% probability that the college graduate has between $14,200 and $33,950 in student loan debt.

Considering the symmetry of the normal distribution, the middle 10% is between the 45th percentile(X when Z = -0.127) and the 55th percentile(X when Z = 0.127), hence:

Z = \frac{X - \mu}{\sigma}

-0.127 = \frac{X - 25150}{12050}

X - 25150 = -0.127(12050)

X = $23,519.65.

Z = \frac{X - \mu}{\sigma}

0.127 = \frac{X - 25150}{12050}

X - 25150 = 0.127(12050)

X = $26,580.35.

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1

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2 years ago
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