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docker41 [41]
3 years ago
11

Please answer this correctly

Mathematics
2 answers:
Naya [18.7K]3 years ago
4 0

Answer:

40-59: 6

60-79: 5

Step-by-step explanation:

If you just added up, you can find all the values.

r-ruslan [8.4K]3 years ago
3 0

Answer:

40 - 59 ⇒ 6

60 - 79 ⇒ 5

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An integer N is to be selected at random from {1, 2, ... , (10)3 } in the sense that each integer has the same probability of be
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Answer:

Probability of N Divisible by 3 - 0.33

Probability of N Divisible by 5 - 0.2

Probability of N Divisible by 7 - 0.413

Probability of N Divisible by 15 - 0.066

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Step-by-step explanation:

Given data:

Integer N {1,2,.....10^3}

Thus total number of ways by which 1000 is divisible by 3 i.e. 1000/3 = 333.3

Probability of N divisible by 3 {N%3 = 0 } = \frac{333.3}{1000} = 0.33

total number of ways by which 1000 is divisible by 5 i.e. 1000/5 = 200

Probability of N divisible by 5 {N%5 = 0 } = \frac{200}{1000} = 0.2

total number of ways by which 1000 is divisible by 7 i.e. 1000/7 = 142.857

Probability of N divisible by 7 {N%7 = 0 } = \frac{142.857}{1000} = 0.413

total number of ways by which 1000 is divisible by 15 i.e. 1000/15 = 66.667

Probability of N divisible by 15 {N%15 = 0 } = \frac{66.667}{1000} = 0.066

total number of ways by which 1000 is divisible by 105 i.e. 1000/105 = 9.52

Probability of N divisible by 105 {N%105 = 0 } = \frac{9.52}{1000} = 0.0095

similarly for N is selected from 1,2.....(10)^k where K is  large then the N value. Therefore effect of k will remain same as previous part.

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3 years ago
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Answer:

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