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Anton [14]
2 years ago
15

A camera manufacturer spends $2,250 each day for overhead expenses plus $6 per camera for labor and materials. The cameras sell

for $16 each. How many cameras must the company sell in one day to equal its daily costs? If the manufacturer can increase production by 50 cameras per day, what would their daily profit be?
Mathematics
2 answers:
andriy [413]2 years ago
8 0
2250 + 6x = 16x
2250 = 16x - 6x
2250 = 10x
2250/10 = x
225 = x <=== they must sell 225 cameras to equal daily costs

if they increase production by 50.......x = 225 + 50 = 275
16(275) - 6(275) - 2250 = 4400 - 1650 - 2250 = 4400 - 3900 = 500.
Their daily profit would be $500.
Jet001 [13]2 years ago
8 0

Answer:

225 cameras

$500 profit

Step-by-step explanation:

The daily cost for overhead expenses is $2250 plus $6 for labor and materials:

Let the total amount spent per day be y:

=2250+6x=y (1)

If the camera sells for $16 we must determine how many must be sold to make 2250+6x. We can write another expression in terms of x and y:

y=16*x  (2)

We can substitute equation 2 into equation 1 and solve x:

16*x=2250+6*x

x=225

The company must sell 225 cameras at a price of $16

If the manafacturer increases the production by 50 a day the daily profit would be:

(225+50)*16=4400

The the profit is the amount made subtracted from the the daily expenses:

4400-(2250+6*275)=500

The daily profit will be $500

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As can be read from your statement written "T, equals, start fraction, left parenthesis, d, minus, 15, right parenthesis, strt superscript, 2, end superscript, divided by, 300, end fraction, plus, 20", I hope your model equation is this :

T = \frac{(d-15)^{2}}{300}  + 20

Hope this is your question, if not I think you will, still be able to find an answer of your question based on this solution

As we have to find lowest average temperature, So for minimum of a function its derivative is equal to 0 there.

So lets find derivative of T function first

So first expand (d-15)^{2} as (d-15)(d-15)

we will use FOIL to multiply these

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so we have T = \frac{d^{2} -30d +225}{300} +20

Now we will derivate each term here,300 in denominator is constant so that will come as it in in denominator.

To derivate terms in dx^{2} -30d +225 we will use power rule formula:

(x^{n} )'= nx^{n-1}

so derivative of (d^{2} )'= 2d^{2-1} = 2d^{1} = 2d

Then derivative of d will be 1

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T' = \frac{2d-30}{300}

For minimum we will put this derivative =0

0 = \frac{2d-30}{300}

Now solve for d

times both sides by 300

0 \times 300 = \frac{2d-30}{300} \times 300

0 = 2d-30

0 +30 = 2d -30 +30

30 = 2d

\frac{30}{2} = \frac{2d}{2}

15 = d

So now we have to find value of lowest temperature.

For that simply plug 15 in d place in original T function equation

T = \frac{(d-15)^{2}}{300}  + 20

T = \frac{(15-15)^{2}}{300}  + 20

T = 20

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2 years ago
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BaLLatris [955]

Answer:

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Step-by-step explanation:

Given

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Required:

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The volume of a cone is calculated as thus;

Volume, V = \frac{1}{3}\pi r^2h

Where V, r and h represent the volume, the radius and the height of the cone respectively

But first we need to calculate the radius of the cone.

Using formula of circumference.

C = 2\pi r

We have C to be 5.8cm

By substituting this value;

5.8 = 2\pi r

Divide through by 2\pi

\frac{5.8}{2\pi} = \frac{2\pi r}{2\pi}

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By substituting r = \frac{2.9}{\pi} and h = 12.5 cm;

V = \frac{1}{3}\pi r^2h becomes

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V = \frac{1}{3}\pi * \frac{8.41}{\pi^2} * 12.5

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Take \pi as 3.14

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According to the chart below, what percentage of Ralph’s expenses are items other than taxes?
pochemuha

Answer:

  • <u>65%</u>

Explanation:

Please, see attached the chart corresponding to this question.

The<em> chart</em> is a pie chart that shows the percentage composition of the <em>expenses </em>and uses color to identify each item.

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