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The two leading digits of each 4-digit number are not shown which number is divisible by 4?
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If two twelve side fair dice are rolled what it the probability that the total number of spots shown is equal to 5?
If two twelve side fair dice are rolled, there are 144 possible outcomes. Of those 144 outcomes, there are four (1-4, 2-3, 3-2, and 4-1) that add up to five, So the probability of rolling a sum of five is 4 in 144, or 1 in 36.
When two dice are rolled what is the probability that both will be the same number?
6 sides on each dice. Each number on the sides is different, this represents the 1 out of all 6 amounts shown. So this is what you'll do for each dice... 1/6 + 1/6 = 2/6 = 1/3
If two fair dice are rolled what is the probability that the total number of spots shown is equal to 6?
P(6) = 5/36 = 0.138888... ≈ 13.9%When two fair dice are rolled, the sample space (all possible outcomes) has 36equiprobable events:(1,1), (1,2), (1,3), (1,4), (1,5),(1,6), (2,1), (2,2), (2,3), (2,4),(2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5),(4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6).There are 5 events where the sum of the dots of each die is equal to 6:(1,5), (2,4), (3,3), (4,2), (5,1).So in two fair dice are rolled the probability that the total number of spots shown is equal to 6 is, P(6) = 5/36 = 0.138888... ≈ 13.9%
What 3 ways probability can be shown?
a:b, a/b, a to b
If two fair dice are rolled what is the probability that the total number of spots shown is equal to 15?
If two fair dice were rolled, there would be 36 outcomes. (1,1),(1,2),....,(6,6) The maximum sum would be 12. Therefore, the probability that the total number of spots shown is equal to 15 is zero.
If two fair dice are rolled what is the probability that the total number of spots shown is equal to ten?
6 & 4 And 5 & 5 are the only two possibilities, so 4 chances in 36 or 1 in nine.
The two leading digits of each 4-digit number are not shown which number is divisible by 4?
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If two twelve side fair dice are rolled what it the probability that the total number of spots shown is equal to 5?
If two twelve side fair dice are rolled, there are 144 possible outcomes. Of those 144 outcomes, there are four (1-4, 2-3, 3-2, and 4-1) that add up to five, So the probability of rolling a sum of five is 4 in 144, or 1 in 36.
When two dice are rolled what is the probability that both will be the same number?
6 sides on each dice. Each number on the sides is different, this represents the 1 out of all 6 amounts shown. So this is what you'll do for each dice... 1/6 + 1/6 = 2/6 = 1/3
When reboarding a PWC after a fall should it be rolled?
It should be rolled in the direction as shown in the manual.
The sum of 3 consective numbers is divisible by 3?
Any three consecutive integers are divisible by three because it can be shown that the sum divided by three is the middle number.
Is 234 divisible by 3 6 and 9?
Using the tests for divisibility:Divisible by 3:Add the digits and if the sum is divisible by 3, so is the original number: 2 + 3 + 4 = 9 which is divisible by 3, so 234 is divisible by 3Divisible by 6:Number is divisible by 2 and 3: Divisible by 2:If the number is even (last digit divisible by 2), then the whole number is divisible by 2. 234 is even so 234 is divisible by 2.Divisible by 3:Already shown above to be divisible by 3. 234 is divisible by both 2 & 3 so 234 is divisible by 6Divisible by 9:Add the digits and if the sum is divisible by 9, so is the original number: 2 + 3 + 4 = 9 which is divisible by 9, so 234 is divisible by 9Thus 234 is divisible by all 3, 6 & 9.
If two fair dice are rolled what is the probability that the total number of spots shown is equal to 6?
P(6) = 5/36 = 0.138888... ≈ 13.9%When two fair dice are rolled, the sample space (all possible outcomes) has 36equiprobable events:(1,1), (1,2), (1,3), (1,4), (1,5),(1,6), (2,1), (2,2), (2,3), (2,4),(2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5),(4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6).There are 5 events where the sum of the dots of each die is equal to 6:(1,5), (2,4), (3,3), (4,2), (5,1).So in two fair dice are rolled the probability that the total number of spots shown is equal to 6 is, P(6) = 5/36 = 0.138888... ≈ 13.9%
When re-boarding a pwc how should it be rolled?
in the direction shown in the manual
When re boarding pwc how should it be rolled?
Direction shown in manual
If two fair dice are rolled what is the probability that the total number of spots shown is equal to 5?
Each of the dice has 6 sides. There are 6 x 6 = 36 possible combinations. Of those, the following equal 5: 1 and 4 2 and 3 3 and 2 4 and 1 So the odds of rolling a 5 is 4 chances out of 36 4 / 36 = 1 / 9 That converts to a probability of 0.1111111...
