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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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3 years ago
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stiks02 [169]
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3 years ago
insurance company checks police records on 559 accidents selected at random and notes that teenagers were at the wheel in 91 of
ss7ja [257]

Answer: (13.22\%,\ 19.34\%)

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Given : Sample space : n= 559

Sample proportion : \hat{p}=\dfrac{91}{559}=0.162790697674\approx0.1628

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Significance level : \alpha= 1-0.95=0.05

Critical value : z_{\alpha/2}=1.96

Confidence level for population proportion:-

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Hence, 95% confidence interval for the percentage of all auto accidents that involve teenage drivers.= (13.22\%,\ 19.34\%)

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faltersainse [42]

Answer:

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I see 4 times 6c, which is,

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Hope this helps!!

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3 years ago
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