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Harman [31]
3 years ago
14

Please help. I've been stuck on this for 3 days. I just really need help. ​

Mathematics
1 answer:
Marysya12 [62]3 years ago
5 0

Answer:

71.5

Step-by-step explanation:

x+1 so (1+10) x (21+5)= 286

but then 1+1=2 and 2x2= 4 so 286 divided by 4= 71.5

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OverLord2011 [107]

Answer: 0

Step-by-step explanation:

The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened.

In this case, event A is the Captain hitting the pirate ship and event B is the pirate hitting the Captain's ship.

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3 years ago
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Are you good at math then here
Nikolay [14]

Answer:

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7 0
3 years ago
The base and sides will cost $.01 per cm2 to produce but the top, which is plastic and resealable, will cost $.02 per cm2 to pro
oksian1 [2.3K]

Here is the full question

A snack food company wishes to have a cylinder package for it's almond and cashew mix. The cylinder must contain 120 cm³ worth of product. The base and sides will cost $.01 per cm2 to produce but the top, which is plastic and resealable, will cost $.02 per cm2 to produce. What should the dimensions be to minimize cost?

Answer:

The radius and height are both dimension in the cylinder; in order to minimize the cost

radius = 2.515 cm

height = 18.93 cm

Step-by-step explanation:

We denote the radius of the cylinder to be = r

and the height of the cylinder = h

The volume of a cylinder is known to be = πr²h

Also, from the question; we are also told that the cylinder contains 120 πcm³

i.e πr²h = 120π

Dividing both sides with π; we have:

r²h = 120

h = \frac{120}{r^2}

The base and sides will cost $.01 per cm² to produce

Total cost of the base and side c_1 = 0.01 ( πr² + 2πrh)

but the top, which is plastic and resealable, will cost $.02 per cm² to produce.

i.e

cost of the top cylinder c_2 = 0.02 ( πr²)

Overall Total cost = c_1 + c_2

= 0.01 ( πr² + 2πrh) + 0.02 ( πr²)

= 0.01 πr² + 0.02 πrh + 0.02 πr²

= 0.03 \pi r^2 + 0.02 \pi r (\frac{120}{r^2} )

= 0.03 \pi r^2 + 2.4 \frac{\pi}{r}

Taking the differentiation to find the radius dimension to minimize cost; we have:

\frac{dc}{dr} =0 ⇒ 0.06 \pi r^2 - \frac{2.4 \pi}{r^2} =0

0.06 \pi r^2 = \frac{2.4 \pi}{r^2}

r^4 = \frac{2.4 \pi}{0.06 \pi}

r^4 = 40

r = \sqrt[4]{40}

r= 2.515 cm

However, \frac{d^2c}{dr^2}=0.12 \pi r + \frac{4.8 \pi}{r^3}

\frac{d^2c}{dr^2}|__{r= 2.515}} =0.12 \pi (2.515) + \frac{4.8 \pi}{(2.515)^3} >0

Therefore; we can say that the cost is minimum at r = 2.515 since it is positive.

To determine the height ; we have:

h = \frac{120}{r^2} \\h = \frac{120}{(2.515)^2}

h = \frac{120}{6.34}

h = 18.93 cm

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