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Mary had $156.00 and spent 1/6 of it for a coat. She then bought a skirt with 1/4 the money she had left. How much money did she

marusya05 [52]

Answer: $ 97.5

Step-by-step explanation:

156 * 1/6 = 26

156-26=130

130 * 1/4- 32.5

130-32.4= 97.5

3 0

2 years ago

In ΔDEF, the measure of ∠F=90°, the measure of ∠D=66°, and EF = 32 feet. Find the length of FD to the nearest tenth of a foot.

Butoxors [25]

Answer:

14.2 ft

Step-by-step explanation:

tan 66° = 32/ FD

FD = 32 / tan 66° = 32 / 2.246 = 14.2

7 0

3 years ago

How is 0.0000267 written in scientific notation

Nataly_w [17]

Answer: 6.7*10^-5

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

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:

0. Write the monthly cost and price-demand equations correctly:

Cost:

$C(x)=72,000+70x$

Price-demand:

$p(x)=300-\dfrac{x}{20}$

Domain:

$0\leq x\leq 6000$

1. Part (A) Find the maximum revenue

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

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. 

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.

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?

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: