648 is the three-digit positive integers have three different digits
According to the statement
we have given that there are three positive digit number are formed with three distinct digits.
And we have to find that the how many words are formed with distinct numbers.
So, to solve this type of problem the Combination formula is best.
Because it provides the all possibilities that from how many ways numbers are formed.
So, from a combination formula
here we take 9 two times because first time when we let a number then remaining numbers are 9. and second time remaining numbers are also 9 because we let the distinct number but for third number there will be a probability that choosing number will be same.
So, Three digit positive number become from 9*9*8 =648
So, 648 is the three-digit positive integers have three different digits.
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Answer:
The 95% confidence interval for the mean is between 0.985g/cm² and 1.047 g/cm².
Step-by-step explanation:
We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
![\alpha = \frac{1-0.95}{2} = 0.025](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B1-0.95%7D%7B2%7D%20%3D%200.025)
Now, we have to find z in the Ztable as such z has a pvalue of
.
So it is z with a pvalue of
, so ![z = 1.96](https://tex.z-dn.net/?f=z%20%3D%201.96)
Now, find M as such
![M = z*\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%2A%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In which
is the standard deviation of the population and n is the size of the sample.
![M = 1.96*\frac{0.155}{\sqrt{94}} = 0.0310](https://tex.z-dn.net/?f=M%20%3D%201.96%2A%5Cfrac%7B0.155%7D%7B%5Csqrt%7B94%7D%7D%20%3D%200.0310)
The lower end of the interval is the mean subtracted by M. So it is 1.016 - 0.0310 = 0.985 g/cm²
The upper end of the interval is the mean added to M. So it is 1.016 + 0.0310 = 1.047 g/cm²
The 95% confidence interval for the mean is between 0.985g/cm² and 1.047 g/cm².
Conditional probablility P(A/B) = P(A and B) / P(B). Here, A is sum of two dice being greater than or equal to 9 and B is at least one of the dice showing 6. Number of ways two dice faces can sum up to 9 = (3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6) = 10 ways. Number of ways that at least one of the dice must show 6 = (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6), (6, 5), (6, 4), (6, 3), (6, 2), (6, 1) = 11 ways. Number of ways of rolling a number greater than or equal to 9 and at least one of the dice showing 6 = (3, 6), (4, 6), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6) = 7 ways. Probability of rolling a number greater than or equal to 9 given that at least one of the dice must show a 6 = 7 / 11
Answer:
x=6
Step-by-step explanation:
3/2=x/4 ( you cross-multiply)
2x=12 ( divide by 2)
x=6
Answer:
7
Step-by-step explanation:
I like to work liter volume problems in terms of decimeters. 1 dm = 10 cm, and 1 dm³ = 1 L, so the tank is 3 dm by 4 dm by 3 dm = 3×4×3 dm³ = 36 L.
36 liters will support 36/5 = 7.2 platies.
7 platies can be kept in the aquarium.