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kiruha [24]
3 years ago
5

A code consists of two digit numbers chosen from 00-99 (i.E 00, 01, 02…..99) followed by two different letters of the alphabet.

What is the probability that the code is 43MZ?
Mathematics
1 answer:
Ede4ka [16]3 years ago
4 0

Answer: 1/67600

Step-by-step explanation:

probability that the code is 43MZ:

Probability = (Total Required outcome / total possible outcomes)

Possible outcomes (digit) = (00,.............. ..., 99)

2.) possible outcomes (alphabet) = (A, B,............... .., Z)

P(43MZ) = P(43) × P(M) × P(Z)

P(43) = 1/100

P(M) = 1/26

P(Z) = 1/26

THEREFORE ;

P(43MZ) = (1/100) × (1/26) × (1/26) = 1/67600

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Convert 0.0045 from standard form to scientific notation.
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A coin is tossed 5 times. Find the probability that all are heads. Find the probability that at most 2 are heads.
fenix001 [56]

Answer:

1/32

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Step-by-step explanation:

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For part B, it is easier to just list the possible outcomes for

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Answer:

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