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1 year ago

9

A local non profit organization is selling popcorn to raise money for hurricane relief. The organization paid $4 per bag for the

popcorn and sold it for $5 per bag. What was the percent markup on each bag of popcorn?

1 answer:

Yuri [45]1 year ago

8 0

Answer: 25%

Step-by-step explanation:

markup or gain is 5-4 = $1

% markup = 1/4 *100

=25℅

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A toy factory makes 5,000 teddy bears per day. The supervisor randomly selects 10 teddy bears from all 5,000 teddy bears and use

tankabanditka [31]

Answer:

9

Step-by-step explanation:

<em>Degrees of freedom</em> is the number of values in the final calculation of a statistic that are free to vary. Degrees of freedom are related to sample size and calculated by n-1 where n is the <em>sample size</em>.

The Supervisor selects 10 teddy bears as sample from 5000 teddy bears produced daily. Therefore in this situation, there are 10-1=9 degrees of freedom

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1 year ago

If the price of attending Big Benefits University is $42,000 a year, including tuition, fees, books, and foregone earnings, what

Furkat [3]

Answer:

$229673.215

Step-by-step explanation:

Given : The price of attending Big Benefits University is $42,000 a year, including tuition, fees, books, and foregone earnings

To Find : what is the marginal cost of attending, if it takes you 5 years to graduate, and you assume a 3% annual inflation rate?

Solution:

Principal = $42000

Time = 5 years

Rate = 3% = 0.03

Formula : $A=P(1+r)^t$

$A_1=42000(1+0.03)^1$

$A_1=43260$

$A_2=42000(1+0.03)^2$

$A_2=44557.8$

$A_3=42000(1+0.03)^3$

$A_3=45894.534$

$A_4=42000(1+0.03)^4$

$A_4=47271.370$

$A_5=42000(1+0.03)^4$

$A_5=48689.511$

So , Marginal cost = 43260+44557.8+45894.534+47271.370+48689.511

Marginal cost = $229673.215

Hence the marginal cost of attending, if it takes you 5 years to graduate, and you assume a 3% annual inflation rate is $229673.215

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1 year ago

Charles uses a graphing calculator to find a quadratic regression model f for a given set of data. When he compares model f to a

makvit [3.9K]

B I hope this helped I’m not sure

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1 year ago

Determine the area (in units2) of the region between the two curves by integrating over the x-axis. y = x2 − 24 and y = 1

astra-53 [7]

Answer:

The area of the region between the two curves by integration over the x-axis is 9.9 square units.

Step-by-step explanation:

This case represents a definite integral, in which lower and upper limits are needed, which corresponds to the points where both intersect each other. That is:

$x^{2} - 24 = 1$

Given that resulting expression is a second order polynomial of the form $x^{2} - a^{2}$ , there are two real and distinct solutions. Roots of the expression are:

$x_{1} = -5$ and $x_{2} = 5$ .

Now, it is also required to determine which part of the interval $(x_{1}, x_{2})$ is equal to a number greater than zero (positive). That is:

$x^{2} - 24 > 0$

$x^{2} > 24$

$x < -4.899$ and $x > 4.899$ .

Therefore, exists two sub-intervals: $[-5, -4.899]$ and $\left[4.899,5\right]$ . Besides, $x^{2} - 24 > y = 1$ in each sub-interval. The definite integral of the region between the two curves over the x-axis is:

$A = \int\limits^{-4.899}_{-5} [{1 - (x^{2}-24)]} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} [{1 - (x^{2}-24)]} \, dx$

$A = \int\limits^{-4.899}_{-5} {25-x^{2}} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} {25-x^{2}} \, dx$

$A = 25\cdot x \right \left|\limits_{-5}^{-4.899} -\frac{1}{3}\cdot x^{3}\left|\limits_{-5}^{-4.899} + x\left|\limits_{-4.899}^{4.899} + 25\cdot x \right \left|\limits_{4.899}^{5} -\frac{1}{3}\cdot x^{3}\left|\limits_{4.899}^{5}$

$A = 2.525 -2.474+9.798 + 2.525 - 2.474$

$A = 9.9\,units^{2}$

The area of the region between the two curves by integration over the x-axis is 9.9 square units.

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A food truck sells sandwiches and

anyanavicka [17]

the answer: is $28 dollars

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