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Ronch [10]
3 years ago
7

PLEASE HELPPPPP PLEASE ILL GIVE YOU BRAINLIST

Mathematics
1 answer:
schepotkina [342]3 years ago
5 0

Answer:

12

Step-by-step explanation:

The number of combinations for  2 shirts 3 pants and 2 shoes can be counted from the tree.

Count the different items at the end of the tree

There are 12 different combinations

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WILL MARK BRILLIANT!!! I need help ASAP
Rashid [163]

Answer:

Step-by-step explanation:

Charging by the quarter mile is for purpose of making that particular taxi service seem cheaper than the others when they post a per mile charge.  If this taxi company is charging .50 per 1/4 mile, they are charging $2 per mile.  So we will base our equation on the per mile charge, not the per quarter-mile charge.  If x is the number of mile driven (our uknown), and we have a flat fee of $2.50 regardless of how many miles we are driven, the cost function in terms of miles is

C(x) = 2x + 2.50

If we are driven 5 miles, then

C(5) = 2(5) + 2.50 so

C(5) = 10 + 2.50 and

C(5) = $12.50

It would cost $12.50 to be driven 5 miles

5 0
3 years ago
A company manufactures and sells x television sets per month. The monthly cost and​ price-demand equations are ​C(x)equals72 com
solmaris [256]

Answer:

Part (A)

  • 1. Maximum revenue: $450,000

Part (B)

  • 2. Maximum protit: $192,500
  • 3. Production level: 2,300 television sets
  • 4. Price: $185 per television set

Part (C)

  • 5. Number of sets: 2,260 television sets.
  • 6. Maximum profit: $183,800
  • 7. Price: $187 per television set.

Explanation:

<u>0. Write the monthly cost and​ price-demand equations correctly:</u>

Cost:

      C(x)=72,000+70x

Price-demand:

     

      p(x)=300-\dfrac{x}{20}

Domain:

        0\leq x\leq 6000

<em>1. Part (A) Find the maximum revenue</em>

Revenue = price × quantity

Revenue = R(x)

           R(x)=\bigg(300-\dfrac{x}{20}\bigg)\cdot x

Simplify

      R(x)=300x-\dfrac{x^2}{20}

A local maximum (or minimum) is reached when the first derivative, R'(x), equals 0.

         R'(x)=300-\dfrac{x}{10}

Solve for R'(x)=0

      300-\dfrac{x}{10}=0

       3000-x=0\\\\x=3000

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):

  • R(3,000) = 300(3,000) - (3000)² / 20 = $450,000

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)

  • Profit (x) = R(x) - C(x)

       Profit(x)=300x-\dfrac{x^2}{20}-\big(72,000+70x\big)

       Profit(x)=230x-\dfrac{x^2}{20}-72,000\\\\\\Profit(x)=-\dfrac{x^2}{20}+230x-72,000

ii) Find the first derivative and equal to 0 (it will be a maximum because the quadratic function is a parabola that opens downward)

  • Profit' (x) = -x/10 + 230
  • -x/10 + 230 = 0
  • -x + 2,300 = 0
  • x = 2,300

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:

  • Profit(2,300) = - (2,300)²/20 + 230(2,300) - 72,000 = 192,500

Hence, the maximum profit is $192,500

iv) Find the price the company should charge for each television set:

Use the price-demand equation:

  • p(x) = 300 - x/20
  • p(2,300) = 300 - 2,300 / 20
  • p(2,300) = 185

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:

  • Profit(x) = -x² / 20 + 230x - 4x - 72,000

  • Profit(x) = -x² / 20 + 226x - 72,000

ii) Find the first derivative and make it equal to 0:

  • Profit'(x) = -x/10 + 226 = 0
  • -x/10 + 226 = 0
  • -x + 2,260 = 0
  • x = 2,260

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:

  • Profit (2,260) = -(2,260)² / 20 + 226(2,260) - 72,000
  • Profit (2,260) = 183,800

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:

  • p(2,260) = 300 - 2,260/20
  • p(2,260) = 187.

That is, the company should charge $187 per television set.

7 0
3 years ago
Which expression has a value that is equivalent to 5n + 13 when n=−5 ?
SSSSS [86.1K]
The answer is C. 4n + 8
8 0
3 years ago
Have no idea how to do it
dalvyx [7]

\bf 2)\\\\ 3+-5\implies 3+(-5)\implies 3+(-3-2)\implies ~~\begin{matrix} 3-3 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~-2\implies -2 \\\\\\ 9)\\\\ -5+(-6)\implies -5+(+5-11)\implies ~~\begin{matrix} -5+5 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~-11\implies -11 \\\\\\ 11)\\\\ 2+(-7)\implies 2+(-2-5)\implies ~~\begin{matrix} 2-2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~-5\implies -5

6 0
4 years ago
The Daily News offers a 15% discount on a Saturday ad that also runs on Tuesday. Skyler Sporting Company contracts for half-page
mart [117]
The ad cost for half page Saturday after discount is

15% of 380 = 0.15×380 = 57
380 - 57 = 323

The ad cost for half page Tuesday after discount is

15% of 330 = 0.15×330 = 49.5
330 - 49.5 = 280.5

Cost of the ads for 10 Saturdays and 10 Tuesdays

(10×323)+(10×280.5) = 3230+2805 = 6035
6 0
3 years ago
Read 2 more answers
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