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MA_775_DIABLO [31]

3 years ago

8

A Fruit Stand sells 6 oranges for $3.00 and grapefruit for $2.40 people buy 10 oranges and 11 grapefruits much does the fruit sp

end

2 answers:

Harlamova29_29 [7]3 years ago

6 0

<span>The <u>correct answer</u> is:

$31.40.

Explanation<span>:

Since 6 oranges cost $3.00, we can divide to find the cost of one orange:

3.00/6 = $0.50. Each orange costs $0.50.

If ten oranges are bought, this would cost 10(0.50)=$5.00.

Each grapefruit costs $2.40; 11 of them would cost 11(2.40) = 26.40.

Together the total cost is $5+$26.40 = $31.40.</span></span>

Over [174]3 years ago

6 0

Answer:

$31.4

Step-by-step explanation:

Given –

Cost of 6 oranges = $3.00

Cost of 1 grapefruit = $2.40

Cost of 1 orange = $3.00/6

= $0.5

People bought 10 oranges and 11 grapefruits.

Cost of 10 oranges = 10 x cost of one orange

= 10 x $0.5

= $ 5

Cost of 11 grapefruit = 11 x cost of one grapefruit

= 11 x $2.4

= $26.4

So total cost of 10 oranges and 11 grapefruits

= $ 5 + $26.4

= $31.4

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An aircraft manufacturer wants to determine the best selling price for a new airplane. The company estimates that the initial co

Blizzard [7]

Answer:

$(a)$ $C(x)=500+20x-5x^{\frac{3}{4}}+0.01x^2$

$p(x)=320-7.7p$

$R(x)=(320-7.7p)p=320p-7.7p^2$

$(b)$ $x=82 \text{planes}$

$(c)$ $p=\$30.91 M\;\; \text{per plane}$

$(d)$ maximum profit $=\$ 15.90M$

Step-by-step explanation:

Given that,

The company estimates that the initial cost of designing the aeroplane and setting up the factories in which to build it will be 500 million dollars.

The additional cost of manufacturing each plane can be modelled by the function.

$m(x)=20x-5x^{\frac{3}{4}}+0.01x^2$

$(a)$ Find the cost, demand (or price), and revenue functions.

$C(x)=500+20x-5x^{\frac{3}{4}}+0.01x^2$

$p(x)=320-7.7p$

$R(x)=(320-7.7p)p=320p-7.7p^2$

$(b)$ Find the production level that maximizes profit.

$f=R(x)-C(x)$

$\Rightarrow f=320p-7.7p^2-(500+20x-5x^{\frac{3}{4}}+0.01x^2)$

$\Rightarrow df=320dp-15.4pdp-20dx+5(\frac{3}{4} )x^{\frac{-1}{4} }dx-0.02xdx$

$x=320-7.7p$

$p=\frac{320-x}{7.7}$

$\frac{dp}{dx} = \frac{-1}{7.7}$

$\frac{df}{dx}=\frac{320}{-7.7} -\frac{15.4(320-x) }{7.7(\frac{-1}{7.7} )}-20+5\frac{3}{4} x^{\frac{-1}{4}} -0.02x=0$

$\Rightarrow -41.5584+83.1169-0.2597x-20+3.75x^{\frac{-1}{4} }-0.02x=0$

$\Rightarrow 21.5585+3.75x^{\frac{-1}{4} }-0.279x=0$

$\Rightarrow x=82 \text{planes}$

$(c)$ Find the associated selling price of the aircraft that maximizes profit.

$p=\frac{320-82}{7.7}$

$\Rightarrow p=\$30.91 M\;\; \text{per plane}$

$(d)$ Find the maximum profit.

Manufacturing cost of one plane is:

$m(1)=20-5+0.01$

$=\$15.01 M$

maximum profit $=\$(30.91-15.01)M$

$=\$15.90M$

3 0

2 years ago

I just need anyone to tell me if it's correct:<br><br> Is 8: True<br> and<br> Is 9: False<br> ?

natta225 [31]

I’m pretty sure

8. would be false because BK would be 12 where as DF is 10.8

9. is true

4 0

3 years ago

Read 2 more answers

find the measure of the vertex anlge of an isosceles triangle if the measure of one base angle is 36°

atroni [7]

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angles in all triangles add up to 180°, so
180°-72°=108°
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3 years ago

What method should I use to solve this?

tekilochka [14]

Answer:

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Step-by-step explanation:

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