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uysha [10]
3 years ago
15

How many 3-digit numbers exist, whose digits are distinct even numbers.

Mathematics
2 answers:
skad [1K]3 years ago
7 0

Answer:

328

Step-by-step explanation:

If z is 0, then x has 9 choices. Therefore, z and x can be chosen in (1 × 9) + (4 × 8) = 41 ways. For each of these ways, y can be chosen in 8 ways. Hence, the desired number is 41 × 8 = 328 numbers 3-digit even numbers exist with no repetitions.

Dahasolnce [82]3 years ago
5 0

Answer:

100

Step-by-step explanation:

my teacher told me and I got it right

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Step-by-step explanation:

Based on the question we are given the percentages of each of the types of candies in the bag except for brown. Since the sum of all the percentages equals 75% and brown is the remaining percent then we can calculate that brown is (100-75 = 25%) 25% of the bag. Now we can show the probabilities of getting a certain type of candy by placing the percentages over the total percentage (100%).

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  • Yellow or Blue: \frac{20}{100} +\frac{20}{100} = \frac{40}{100}  ....add the numerators
  • Not Green:  \frac{80}{100}.... since the sum of all the rest is 80%
  • Stiped:  \frac{25}{100} .... there are 0 striped candies.

Assuming the <u><em>ratios/percentages</em></u> of the candies stay the same having an infinite amount of candy will not affect the probabilities. That being said in order to calculate consecutive probability of getting 3 of a certain type in a row we have to multiply the probabilities together. This is calculated by multiplying the numerators with numerators and denominators with denominators.

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\frac{15*85*15}{100*100*100} = \frac{19,125}{1,000,000} = \frac{1.9125}{100}

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  • At least 1 green: multiply the percent of green by 100% twice, since the other two can by any

\frac{20*100*100}{100*100*100} = \frac{200,000}{1,000,000} = \frac{20}{100}

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