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3 years ago

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As part of a new advertising campaign, a beverage company wants to increase the dimensions of their cans by a multiple of 1.12.

If the cans are currently 12 cm tall, 6 cm in diameter, and have a volume of 339.12 cm3, how much more will the new cans hold? Use 3.14 for π and round your answer to the nearest hundredth.

1 answer:

olga_2 [115]3 years ago

3 0

Answer:

The new can can hold upto volume of 476.69 cubic centimeters.

Step-by-step explanation:

Diameter of the can = d = 6 cm

Radius of the can = r = 0.5d = 0.5 × 6 cm = 3 cm

Height of the can = h = 12 cm

Company wants to increase the dimensions of their cans by a multiple of 1.12. So, the new dimension will be:

r'= 3 cm × 1.12 = 3.36 cm

h' = 12 cm × 1.12 =13.44 cm

Volume of the = can = V

Volume of a cylinder = $\pi r^2h$

$V=\pi r'^2h'$

$=3.14\times (3.36 cm)^2\times 13.44 cm=476.6808 cm^3\approx 476.68 cm^3$

The new can can hold upto volume of 476.69 cubic centimeters.

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REAL LIFE A cylindrical hazardous waste container has a diameter of 1.5 feet and a height of 1.6 feet. Abo

aivan3 [116]

Answer:

It can hold up to $1.256ft^3$ of waste

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Step-by-step explanation:

Given

$D = 1.5ft$ -- Diameter

$h = 1.6ft$ -- height

Required

Determine the amount of waste it can hold

This simply means that we calculate the volume of the cylindrical container.

And this is calculated using:

$Volume = \pi r^2h$

Where

$\pi = 3.14$

and

$r = \frac{D}{2}$

$r = \frac{1.5}{2}$

$r = 0.75$

So, we have:

$Volume = 3.14 * 0.5^2 * 1.6$

$Volume = 1.256ft^3$

<em>Hence, it can hold up to </em> $1.256ft^3$ <em> of waste</em>

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3 years ago

An expression is shown below

Leno4ka [110]

Answer:

B) It is irrational and equal to $= 4\sqrt{2}$

Step-by-step explanation:

The expression, $\sqrt{18} + \sqrt{2}$ , is an irrational expression because it contains two irrational numbers, $\sqrt{18}$ and $\sqrt{2}$ . Both number are not perfect square. there is no whole number that would be squared that will give you 18 nor 2.

The expression can be equals the following:

$\sqrt{18} + \sqrt{2}$

$= \sqrt{9*2} + \sqrt{2}$

$= \sqrt{9}*\sqrt{2} + \sqrt{2}$

$= 3\sqrt{2} + \sqrt{2}$

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Dafna11 [192]

1 question what is x equal to?

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Suppose that 4 fair coins are tossed. Let Equals The event that exactly 2 coins show tails and Equal The event that at least 2 c

faust18 [17]

Answer:

a) P ( E | F ) = 0.54545

b) P ( E | F' ) = 0

Step-by-step explanation:

Given:

- 4 Coins are tossed

- Event E exactly 2 coins shows tail

- Event F at-least two coins show tail

Find:

- Find P ( E | F )

- Find P ( E | F prime )

Solution:

- The probability of head H and tail T = 0.5, and all events are independent

So,

P ( Exactly 2 T ) = ( TTHH ) + ( THHT ) + ( THTH ) + ( HTTH ) + ( HHTT) + ( HTHT) = 6*(1/2)^4 = 0.375

P ( At-least 2 T ) = P ( Exactly 2 T ) + P ( Exactly 3 T ) + P ( Exactly 4 T) = 0.375 + ( HTTT) + (THTT) + (TTHT) + (TTTH) + ( TTTT)

= 0.375 + 5*(1/2)^4 = 0.375 + 0.3125 = 0.6875

- The probabilities for each events are:

P ( E ) = 0.375

P ( F ) = 0.6875

- The Probability to get exactly two tails given that at-least 2 tails were achieved:

P ( E | F ) = P ( E & F ) / P ( F )

P ( E | F ) = 0.375 / 0.6875

P ( E | F ) = 0.54545

- The Probability to get exactly two tails given that less than 2 tails were achieved:

P ( E | F' ) = P ( E & F' ) / P ( F )

P ( E | F' ) = 0 / 0.6875

P ( E | F' ) = 0

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const2013 [10]

Answer:

Let us assume the total number of students surveyed = x

Then

Number of students preferring red = x/4

Number of students preferring blue = x/8

Then

Number of students remaining = x - [(x/4) + (x/8)]

=x - [(2x + x)/8)

= x - (3x/8)

= (8x - 3x)/8

= 5x/8

Number of students preferring the color green = (3/5) * (5x/8)

= 3x/8

We also know that the total number of students preferring the color green is 15

So

3x/8 = 15

3x = 15 * 8

x = (15 * 8)/3

= 5 * 8

= 40

So in total 40 students were surveyed. I hope the procedure is clear to you.

Step-by-step explanation:

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