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almond37 [142]
3 years ago
13

A division of a company produces income tax apps for smartphones. Each income tax app sells for $9. The monthly fixed costs incu

rred by the division are $20,000, and the variable cost of producing each income tax app is $4.
(a) Find the break-even point for the division.

(x, y) =



Incorrect: Your answer is incorrect.





(b) What should be the level of sales in order for the division to realize a 10% profit over the cost of making the income tax apps? (Round your answer up to the nearest whole number.)

income tax apps

Mathematics
1 answer:
qaws [65]3 years ago
7 0

Answer:

a) The break-even point for the division (x, y) = ( 4000, 36000)

b) ) What should be the level of sales in order for the division to realize a 10% profit = 4783

Step-by-step explanation:

The detailed steps and appropriate analysis is as shown in the attachment.

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A student is trying to solve the system of two equations given below: Equation P: a + b = 6 Equation Q: 4a + 2b = 19 Which of th
Marrrta [24]

Answer:

-4(a + b = 6)

Step-by-step explanation:

Given

a + b = 6

4a + 2b = 19

Required

Eliminate a

Multiply the first equation by -4

-4(a + b = 6)

Add to the second equation

-4(a + b = 6) + (4a + 2b = 19)

Solve brackets

(-4a -4b = -24) + (4a + 2b = 19)

Open bracket

-4a + 4a -4b  + 2b = -24 + 19

-4b  + 2b = -24 + 19

At this point, a has been eliminated;

From the list of given options, the option that answers the question is -4(a + b = 6)

8 0
4 years ago
A lake polluted by bacteria is treated with an antibacterial chemical. Aftertdays, thenumber N of bacteria per milliliter of wat
jeyben [28]

Answer:

N(1)=50 is a minimum

N(15)=4391.7 is a maximum

Step-by-step explanation:

<u>Extrema values of functions </u>

If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.  

We are given a function (corrected)

N(t) = 20(t^2-lnt^2)+ 30

N(t) = 20(t^2-2lnt)+ 30

(a)

First, we take its derivative

N'(t) = 20(2t-\frac{2}{t})

Solve N'(t)=0

20(2t-\frac{2}{t})=0

Simplifying

2t^2-2=0

Solving for t

t=1\ ,t=-1

Only t=1 belongs to the valid interval 1\leqslant t\leqslant 15

Taking the second derivative

N''(t) = 20(2+\frac{2}{t^2})

Which is always positive, so t=1 is a minimum

(b)

N(1)=20(1^2-2ln1)+ 30

N(1)=50 is a minimum

(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15

(d)

N(15)=20(15^2-2ln15)+ 30

N(15)=4391.7 is a maximum

7 0
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11Alexandr11 [23.1K]

9514 1404 393

Answer:

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Step-by-step explanation:

The change is found by subtracting the start temperature from the finish temperature. It is ....

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