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torisob [31]
3 years ago
6

how much material was used in the manufacture of 24000 celluloid dice,if each dice has an edge of 1/4 inches?

Mathematics
2 answers:
sasho [114]3 years ago
8 0
The volume of each celluloid die is (.25 x .25 x .25) = 0.015625 cubic inch.

To manufacture 24,000 of them, you need to start with <u>at least</u>

(24,000) x (0.015625) = <u>375 cubic inches</u>.

I don't know how celluloid is sold, so you should also keep in mind that
375 cubic inches = about 207.8 fluid ounces, or about 6.5 quarts.

I'm sure a bit more than that was used in the manufacture, since
there's always some wasted, spilled, or trimmed off of the edges.


Anika [276]3 years ago
4 0
Material was used in the manufacture of 24000 celluloid dice
 each dice has an edge of 1/4 inches
=1/4*24000
=24000/4
=6000
material was used in the manufacture of 24000 celluloid dice is 6000
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mamaluj [8]
<h2>Answer:</h2>

(a)

The probability is :  1/2

(b)

The probability is :  1/2

<h2>Step-by-step explanation:</h2>

The numbers​ 1, 2,​ 3, 4, and 5 are written on slips of​ paper, and 2 slips are drawn at random one at a time without replacement.

The total combinations that are possible are:

(1,2)   (1,3)    (1,4)    (1,5)

(2,1)   (2,3)   (2,4)   (2,5)

(3,1)   (3,2)   (3,4)   (3,5)

(4,1)   (4,2)   (4,3)   (4,5)

(5,1)   (5,2)   (5,3)   (5,4)

i.e. the total outcomes are : 20

(a)

Let A denote the event that the first number is 4.

and B denote the event that the sum is: 9.

Let P denote the probability of an event.

We are asked to find:

               P(A|B)

We know that it could be calculated by using the formula:

P(A|B)=\dfrac{P(A\bigcap B)}{P(B)}

Hence, based on the data we have:

P(A\bigcap B)=\dfrac{1}{20}

( Since, out of a total of 20 outcomes there is just one outcome which comes in A∩B and it is:  (4,5) )

and

P(B)=\dfrac{2}{20}

( since, there are just two outcomes such that the sum is: 9

(4,5) and (5,4) )

Hence, we have:

P(A|B)=\dfrac{\dfrac{1}{20}}{\dfrac{2}{20}}\\\\i.e.\\\\P(A|B)=\dfrac{1}{2}

(b)

Let A denote the event that the first number is 3.

and B denote the event that the sum is: 8.

Let P denote the probability of an event.

We are asked to find:

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Hence, based on the data we have:

P(A\bigcap B)=\dfrac{1}{20}

( since, the only outcome out of 20 outcomes is:  (3,5) )

and

P(B)=\dfrac{2}{20}

( since, there are just two outcomes such that the sum is: 8

(3,5) and (5,3) )

Hence, we have:

P(A|B)=\dfrac{\dfrac{1}{20}}{\dfrac{2}{20}}\\\\i.e.\\\\P(A|B)=\dfrac{1}{2}

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