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3 years ago

Solve P=R-C=XP-C P=140-0.001x C=40x/150,000

1 answer:

7 0

X=quantity soldunit Price,P=140-0.001xunit Cost, C=40x/150000total Revenue, R=xPProfit, Pr where
Pr=R-C=xP-C=x(140-0.001x)-40x/150000=140x-0.001x^2-40x/150000
To maximize profit,Pr'=0140-0.002x-40/150000=>x=70,000 units

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Find the average of f(x)=x^3-x over [0,2]

vampirchik [111]

9514 1404 393

Answer:

Step-by-step explanation:

The average value of the function is the integral over the interval, divided by the width of the interval.

$\displaystyle\overline{y}=\frac{1}{2-0}\int_0^2{f(x)}\,dx=\left.\frac{1}{2}\left(\frac{x^4}{4}-\frac{x^2}{2}\right)\right|_0^2=\frac{1}{2}\left(\frac{2^4}{4}-\frac{2^2}{2}\right)=\frac{4-2}{2}\\\\\boxed{\overline{y}=1}$

3 0

3 years ago

Gisselle has 1/2 of a candy bar. She role off 1/2 of what she had. She then took the piece she broke off and split it into three

saul85 [17]

Answer:

$\frac{1}{12}$

Step-by-step explanation:

Given:

Gisselle has 1/2 of a candy bar.

She role off 1/2 of what she had and split it into three equal pieces.

Question asked:

What fraction of a candy bar was each piece?

Solution:

Gisselle has = $\frac{1}{2} \ of \ a \ candy \ bar$

She role off = $\frac{1}{2} \ of \ what\ she \ had$

$=\frac{1}{2} \ of\ \frac{1}{2}\\=\frac{1}{2} \times \frac{1}{2}\\\\=\frac{1}{4}$

As she split it into three equal pieces, we will divided it by 3, we get,

$\frac{1}{4} \div3\\\frac{1}{4}\times\frac{1}{3}$

$\frac{1}{12}$

Therefore, fraction of a candy bar of each piece is $\frac{1}{12}$

4 0

3 years ago

Julie went to the 10% off store and bought 3 candy bars for $1.30 each, a loaf of bread for $2.90, and 2 cases of Dr. Pepper for

uysha [10]

You will save $1.98 cents , :)

3 0

3 years ago

I need help, it’s geometry:

VashaNatasha [74]

Answer: The area is 5x + 20

Step-by-step explanation:

Add all the numbers up.

7 + 4x - 8 = 3

13 + (x + 8) = (x + 21)

Therefore, your answer is 5x + 20

Let me know if you have any questions!

4 0

3 years ago

The enrollments of all 13 public universities in the state of ohio are listed below. college enrollment university of akron 25,6

bekas [8.4K]

Answers:

(a) The enrollment data is a population
(b) Mean: 22,150.62
(c) Median: 18,670
(d) Range: 57,776
(e) Standard Deviation: 14,258.49

Explanations:

(a) As we noticed in the problem, the enrollment data is used for the study of the number of enrollments in all of the public universities in Ohio. Since the given enrollment data contains the number of enrollments in all of the public univeristies in Ohio, the enrollment data is a population.

(b) To compute the mean, we get the sum of all the data points and divide it by the number of data points.

Since,

Sum of all data points = 1725 + 4344 + 14655 + 15693 + 17338 + 18641 18670 + 20251 + 20839 + 25612 + 34187 + 36502 + 59,501
Sum of all data points = 287,958

and the number of data points = 13,

mean = 287,958 ÷ 13 = 22,150.62

(c) The median is computed when the data points are arranged from least to greatest and it is equal to either:

>> the middle number - if the number of data points is an odd number, or
>> the average of 2 middle numbers - if the number of data points is an even number

In the enrollment data, when the data points are arranged from lowest to highest, the following is the result:

1725, 4344, 14655, 15693, 17338, 18641, 18670, 20251, 20839, 25612, 34187, 36502, 59,501

Since the number of data points is 13 and 13 is an odd number, we find the middle number, which is the 7th lowest number = 18670.

Hence, the median is 18670.

(d) The range of the data is the difference between the highest data point and the lowest data point.

Since the highest data point is 59,501, the lowest data point is 1725 and their difference is 59,501 - 1725 = 57,776. The range of the enrollment data is 57,776.

(e) To compute for standard deviation,

>> First, we compute the squares of all the data points and find their sum:

Sum of the squares of data points = 1725² + 4344² + 14655² + 15693² + 17338² + 18641² + 18670² + 20251² + 20839² + 25612² + 34187² + 36502² + 59,501²

Sum of the squares of data points = 9,021,404,700

>> Then, we compute the average of the squares of all data points by dividing the sum of the squares of all data points by the number of data points:

Average of the squares
= Sum of the squares of data points ÷ number of data points
= 9,021,404,700 ÷ 13
Average of the squares = 693,954,207.69231
 
>> Finally, the standard deviation is the square root of the difference of the average of squares and the square of the average of the data points:

$\sigma = \sqrt{E(X^2) -(E(X))^2}$

where:

$\sigma = \text{standard deviation}
\\ E(X) = \text{average of all data points} = 22,150.62
\\ E(X^2) = \text{average of the squares of all data points} \approx 693,954,207.69231
$

So, the standard deviation is given by

$\sigma = \sqrt{E(X^2) -(E(X))^2}
\\ \sigma = \sqrt{693,954,207.69231 -(22,150.62)^2}
\\ \boxed{\sigma \approx 14,258.49}
$

Therefore, the standard deviation is 14,258.49.

3 0

3 years ago