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Lesechka [4]
3 years ago
15

How many zeros are in the product of any nonzero whole number less than 10 and 500

Mathematics
1 answer:
worty [1.4K]3 years ago
8 0

Answer:

If the non-zero whole number is even, then there are 3 zeroes

If the non-zero whole number is odd, then there are 2 zeroes

Step-by-step explanation:

* <em>Lets explain the information in the problem</em>

- Whole numbers are positive numbers, including zero, without

 any decimal or fractional

- Nonzero whole numbers are positive numbers, not including zero

- The non-zero whole numbers less then 10 are:

  1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9

- The product of any nonzero whole number less than 10 and 500

  means multiply any of nonzero number above by 500

- The number of zeros in the product is equal to the sum of the

  number of zeros at the end of each of the factors.

# 1 × 500 = 5 × 100 ⇒ 2 zeros

# 2 × 500 = 2 × 5 × 100 = 10 × 100 ⇒ 3 zeros

# 3 × 500 = 3 × 5 × 100 = 15 × 100 ⇒ 2 zeros

# 4 × 500 = 2 × 2 × 5 × 100 = 2 × 10 × 100 ⇒ 3 zeros

# 5 × 500 = 5 × 5 × 100 = 25 × 100 ⇒ 2 zeroes

# 6 × 500 = 2 × 3 × 5 × 100 = 3 × 10 × 100 ⇒ 3 zeros

# 7 × 500 = 7 × 5 × 100 = 35 × 100 ⇒ 2 zeroes

# 8 × 500 = 2 × 2 × 2 × 5 × 100 = 4 × 10 × 100 ⇒ 3 zeros

# 9 × 500 = 3 × 3 × 5 × 100 = 45 × 100 ⇒ 2 zeros

- From all answers above:

# If the non-zero whole number is even (2 , 4 , 6 , 8), then there are

  3 zeros in the product

# If the non-zero whole number is odd (1 , 3 , 5 , 7 , 9), then there are

  2 zeros in the product

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12% of children are nearsighted, but this condition often is not detected until they go to kindergarten. A school district tests
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Answer:

There is a 34.60% probability that 0 or 1 of them is nearsighted.

Step-by-step explanation:

For each children, there are only two possible outcomes. Either they are nearsighted, or they are not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

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And p is the probability of X happening.

In this problem, we have that:

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This probability is:

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