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Natasha_Volkova [10]
3 years ago
13

A prism is completely filled with 2304 cubes that have edge length of 14 cm

Mathematics
2 answers:
boyakko [2]3 years ago
4 0
2304cubes(14^3 cm^3/cube)=6322176 cm^3
mash [69]3 years ago
3 0
54.86.......................................................................
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i belive that the answer is 0.2 because 0.2 is 1/5 of a whole


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Which side lengths form a right triangle ?
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Step-by-step explanation:

1^2 + 8^2 = c^2

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2 years ago
Illinois license plates used to consist of either three letters followed by three digits or two letters followed by four digits.
DENIUS [597]

Answer:

The probability that a randomly chosen plate contains the number 2222 is 0.000028 approximately.

The probability that a randomly chosen plate contains the sub-string HI is 0.002548  approximately.

Step-by-step explanation:

Consider the provided information.

Illinois license plates used to consist of either three letters followed by three digits or two letters followed by four digits.

Part (A)

Let A is the ways in which plates consist of three letters followed by three digits and B is the ways in which two letters followed by four digits.

Here repetition is allow. The number of alphabets are 26 and the number of distinct digits are 10.

The numbers of ways in which three letters followed by three digits can be chosen is: 26\times 26\times 26 \times 10 \times 10 \times10

26^3\times 10^3=17576000

The numbers of ways in which two letters followed by four digits can be chosen is: 26\times 26\times 10 \times 10 \times 10 \times10

26^2\times 10^4=6760000

Hence, the total number of ways are 17576000 + 6760000 = 24336000

Randomly chosen plate contains the number 2222 that means the first two letter can be any alphabets but the rest of the digit should be 2222.

Thus, the total number of ways that a randomly chosen plate contains the number 2222 number are: 26^2=676

The probability that a randomly chosen plate contains the number 2222 is:

\frac{676}{24336000} \approx 0.000028

Part (B)

The number of ways in which chosen plate contains the sub-string HI:

If three letters followed by three digits plate contains the sub-string HI, then the number of possible ways are:

26\times 1\times10^3+1\times 26\times10^3

If two letters followed by four digits plate contains the sub-string HI, then the number of possible ways are:

1\times 10^4

Thus, the total number of ways that a randomly chosen plate contains the sub-string HI are:

1\times 10^4+26\times 1\times10^3+1\times 26\times10^3

62000

From part (A) we know that the total number of ways to chose a number plate is 24336000.

The probability that a randomly chosen plate contains the sub-string HI is:

\frac{62000}{24336000} \approx 0.002548

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N would equal 57 × 3 which is 171.
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