Answer:
<em>(A) 560 students</em>
<em>(B) P = 0.41</em>
Step-by-step explanation:
<em>(Please check the attached picture for more details)</em>
As given, a survey of 1000 students found that:
41% of the surveyed students were freshmen
=> 1000 x 41/100 = 410 surveyed students were freshmen
31% were juniors
=> 1000 x 31/100 = 310 surveyed students were junior
67% of surveyed students lived on the college campus
=> 1000 x 67/100 = 670 surveyed students lived on the campus
<em>(A) 260 of the surveyed students were freshmen living on the campus</em>
=> A = 410 - 260 = 150 surveyed students were freshmen NOT living on the campus
=> B = 670 - 260 = 410 surveyed students were not freshmen living on the campus
=> The number of surveyed students who were either freshmen or living on the campus:
N = A + B = 150 + 410 = 560
<em>(B) From a previous survey you know that of interviewed juniors, only 23% live on campus (we don't really need this)</em>
=>The probability a surveyed student is a junior or someone that lives on campus: P = (670 - 260)/1000 = 0.41
I hope this helps!
For this case we must factor the following expression:
To factor, we must find two numbers that when added together give -7 and when multiplied by -10.
It is observed that there are not two whole numbers that comply with the aforementioned.
Thus, the following equation applies:
Where:
We replace:
So, the roots are:
Answer:
Answer:
Part (A)
Part (B)
Part (C)
Explanation:
<u>0. Write the monthly cost and price-demand equations correctly:</u>
Cost:
Price-demand:
Domain:
<em>1. Part (A) Find the maximum revenue</em>
Revenue = price × quantity
Revenue = R(x)
Simplify
A local maximum (or minimum) is reached when the first derivative, R'(x), equals 0.
Solve for R'(x)=0
Is this a maximum or a minimum? Since the coefficient of the quadratic term of R(x) is negative, it is a parabola that opens downward, meaning that its vertex is a maximum.
Hence, the maximum revenue is obtained when the production level is 3,000 units.
And it is calculated by subsituting x = 3,000 in the equation for R(x):
Hence, the maximum revenue is $450,000
<em>2. Part (B) Find the maximum profit, the production level that will realize the maximum profit, and the price the company should charge for each television set. </em>
i) Profit(x) = Revenue(x) - Cost(x)
ii) Find the first derivative and equal to 0 (it will be a maximum because the quadratic function is a parabola that opens downward)
Thus, the production level that will realize the maximum profit is 2,300 units.
iii) Find the maximum profit.
You must substitute x = 2,300 into the equation for the profit:
Hence, the maximum profit is $192,500
iv) Find the price the company should charge for each television set:
Use the price-demand equation:
Therefore, the company should charge a price os $185 for every television set.
<em>3. Part (C) If the government decides to tax the company $4 for each set it produces, how many sets should the company manufacture each month to maximize its profit? What is the maximum profit? What should the company charge for each set?</em>
i) Now you must subtract the $4 tax for each television set, this is 4x from the profit equation.
The new profit equation will be:
ii) Find the first derivative and make it equal to 0:
Then, the new maximum profit is reached when the production level is 2,260 units.
iii) Find the maximum profit by substituting x = 2,260 into the profit equation:
Hence, the maximum profit, if the government decides to tax the company $4 for each set it produces would be $183,800
iv) Find the price the company should charge for each set.
Substitute the number of units, 2,260, into the equation for the price:
That is, the company should charge $187 per television set.