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Greeley [361]
3 years ago
7

Currently, the demand equation for necklaces is Q = 30 – 4P. The current price is $10 per necklace. Is this the best price to ch

arge in order to maximize revenues?
Mathematics
1 answer:
barxatty [35]3 years ago
6 0

Answer:

The correct answer is NO. The best price to be charged is $3.75

Step-by-step explanation:

Demand equation is given by Q = 30 - 4P, where Q is the quantity of necklaces demanded and P is the price of the necklace.

⇒ 4P = 30 - Q

⇒ P = \frac{30-Q}{4}

The current price of the necklace $10.

Revenue function is given by R = P × Q = \frac{1}{4} × ( 30Q - Q^{2})

To maximize the revenue function we differentiate the function with respect to Q and equate it to zero.

\frac{dR}{dQ} =  \frac{1}{4} × ( 30 - 2Q) = 0

⇒ Q = 15.

The second order derivative is negative showing that the value of Q is maximum.

Therefore P at Q = 15 is $3.75.

Thus to maximize revenue the price should be $3.75.

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When Andrew goes bowling, his scores are normally distributed with a mean of 115 and a standard deviation of 11. What is the pro
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Answer:

I have no idea. : /

Step-by-step explanation:

7 0
3 years ago
Any help would be appreciated. Thank you!
trapecia [35]

Answer:

Area =62.5\sqrt{6} square units

AB=5\sqrt{15} units

BC=5\sqrt{10} units

Step-by-step explanation:

In a right triangle the altitude drawn to the hypotenuse is the geometric mean of the segments at which this altitude divides the hypotenuse.

So,

BD^2=15\cdot 10\\ \\BD^2=150\\ \\BD=\sqrt{150}=5\sqrt{6}\ units

a. The area of the triangle ABC is

A_{ABC}=\dfrac{1}{2}\cdot BD\cdot AC=\dfrac{1}{2}\cdot 5\sqrt{6}\cdot (15+10)=\dfrac{125\sqrt{6}}{2}=62.5\sqrt{6}\ un^2.

b. The legs of the right triangle are geometric means of the segment adjacent to this leg and the hypotenuse, so

AB^2=AD\cdot AC=15\cdot 25\Rightarrow AB=5\sqrt{15}\ units\\ \\BC^2=CD\cdot AC=10\cdot 25\Rightarrow BC=5\sqrt{10}\ units

5 0
3 years ago
What is answer to this question ?
Semenov [28]
A its not a function 
6 0
3 years ago
What is a three digit number thats an odd multiple of three and the product of its digits is 24 and is larger than 15 squared?
Reika [66]
The only way 3 digits can have product 24 is 
1 x 3 x 8 = 241 x 4 x 6 = 242 x 2 x 6 = 242 x 3 x 4 = 24
So the digits comprises of 1,3,8 or 1,4,6, or 2,2,6, or 2,3,4 
To be divisible by 3 the sum of the digits must be divisible by 3.
1+  3+ 8=12, 1+ 4+ 6= 11, 2 +2 + 6=10, 2 +3 + 4=9Of those sums of digits, only 12 and 9 are divisible by 3. 

So we have ruled out all but integers whose digits consist of1,3,8, and 2,3,4.

Meanwhile they must be odd they either must end in 1 or 3.
The only ones which can end in 1 are 381 and 831.

The others must end in 3. 

They must be greater than 152 which is 225. So the

First digit cannot be 1. So the only way its digits can contain of1,3,8 and close in 3 is to be 813.

The rest must contain of the digits 2,3,4, and the only way they can end in 3 is to be 243 or 423.  
So there are precisely five such three-digit integers: 381, 831, 813, 243, and 423.
7 0
3 years ago
You are working for a company that creates tubes that measure 10 cm long and have a diameter of 6 cm. The tubes need to be cut j
Nataly_w [17]

Answer:

A= 2\pi (3cm)^2 +1 2\pi (3cm)(10 cm)= 18 \pi + 60\pi = 78 \pi cm^2

And replacing the value of \pi =3.14 we got:

A = 78*3.14 cm^2= 244.92 cm^2

Step-by-step explanation:

For this case we have the long who represent the height 10 cm and the diameter is 6cm. So then the radius is given by:

r = \frac{D}{2}= \frac{6cm}{2}= 3cm

And the surface area for a cylinder is given by this formula:

A= 2\pi r^2 +2 \pi rh

And replacing the info given we got:

A= 2\pi (3cm)^2 +1 2\pi (3cm)(10 cm)= 18 \pi + 60\pi = 78 \pi cm^2

And replacing the value of \pi =3.14 we got:

A = 78*3.14 cm^2= 244.92 cm^2

4 0
3 years ago
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