How many 3-digit numbers contain exactly one 1?

2 answers:

8 0

Given a 3-digit number containing exactly one "1".
(ASSUME base-10 numbers, so digits are from 0-9)

Number contains exactly one "1", therefore the two other digits are limited to 2,3,4,....9,0.
If the "1" is in the leftmost position, there are 9*9=81 ways to arrange the second and third positions.
If the "1" is in the second or third position, then there are 8 choices for the leftmost digit (2-9), and 9 choices for the other digit for a total of 2*8*9=144 arrangements.
So the total number of 3-digit numbers containing exactly one "1" is 81+144=225.

marysya [2.9K]3 years ago

6 0

I believe the answer would be 17, because 100, 210, 201, 301, 310, 401, 410, 501, 510, 601, 610, 701, 710, 801, 810, 901, 910

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John has two jobs. For daytime work at a jewelry store he is paid

djyliett [7]

Given Information:

John's mean monthly commission = μ = $10,000

Standard deviation of monthly commission = σ = $2,000

Answer:

$P(9,000 < X < 11,000) = 0.383\\\\P(9,000 < X < 11,000) = 38.3 \%$

The probability that John's commission from the jewelry store is between $9,000 and $11,000 is 38.3%

Step-by-step explanation:

What is Normal Distribution?

We are given a Normal Distribution, which is a continuous probability distribution and is symmetrical around the mean. The shape of this distribution is like a bell curve and most of the data is clustered around the mean. The area under this bell shaped curve represents the probability.

We want to find out the probability that John's commission from the jewelry store is between $9,000 and $11,000?

$P(9,000 < X < 11,000) = P( \frac{x - \mu}{\sigma} < Z < \frac{x - \mu}{\sigma} )\\\\P(9,000 < X < 11,000) = P( \frac{9,000 - 10,000}{2,000} < Z < \frac{11,000 - 10,000}{2,000} )\\\\P(9,000 < X < 11,000) = P( \frac{-1,000}{2,000} < Z < \frac{1,000}{2,000} )\\\\P(9,000 < X < 11,000) = P( -0.5 < Z < 0.5 )\\\\P(9,000 < X < 11,000) = P( Z < 0.5 ) - P( Z < -0.5 ) \\\\$

The z-score corresponding to 0.50 is 0.6915

The z-score corresponding to -0.50 is 0.3085

$P(9,000 < X < 11,000) = 0.6915 - 0.3085 \\\\P(9,000 < X < 11,000) = 0.383\\\\P(9,000 < X < 11,000) = 38.3 \%$

Therefore, the probability that John's commission from the jewelry store is between $9,000 and $11,000 is 38.3%

How to use z-table?

Step 1:

In the z-table, find the two-digit number on the left side corresponding to your z-score. (e.g 1.4, 2.2, 0.5 etc.)

Step 2:

Then look up at the top of z-table to find the remaining decimal point in the range of 0.00 to 0.09. (e.g. if you are looking for 0.50 then go for 0.00 column)

Step 3:

Finally, find the corresponding probability from the z-table at the intersection of step 1 and step 2.

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