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Wewaii [24]
3 years ago
12

If you roll two fair six sided dice what's the probability that both dice show an odd number

Mathematics
2 answers:
laila [671]3 years ago
8 0

Answer with explanation:

When a die is thrown once, total possible outcome ={1,2,3,4,5,6}=6

Total number of odd number in total possible outcome when a die is thrown once ={1,3,5}=3

Probability of an event

                                =\frac{\text{Total Possible Outcome}}{\text{Total favorable outcome}}

Probability of getting an Odd number when a Dice is thrown once

         =\frac{3}{6}\\\\=\frac{1}{2}

So, When a two fair six sided dice is rolled, the probability that both dice show an odd number

              =\frac{1}{2}\times \frac{1}{2}\\\\=\frac{1}{4}

seraphim [82]3 years ago
3 0

Answer: \dfrac{1}{4}

Step-by-step explanation:

If we roll  two fair six sided dice, then the total number of possible outcomes :-

6\times6=36    ( by fundamental principle of counting.)

Total Odd numbers  (1,3,5)= 3  

Favorable outcomes for both dice show an odd number = 3\times3=9            ( by fundamental principle of counting.)

Now, the  probability that both dice show an odd number :-

\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}\\\\=\dfrac{9}{36}=\dfrac{1}{4}

Hence, the probability that both dice show an odd number =\dfrac{1}{4}

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\rm \to m =  \dfrac{ - 1 - 3}{4 - 8}

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2 years ago
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jekas [21]

Answer:

B

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here b = 8 and h = 5, so

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3 years ago
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Find the smallest positive integer n such that the digit sum of n is divisible by 5, and the digit sum of n +1 is also divisible
alisha [4.7K]

Answer:

139,999

Step-by-step explanation:

If the digit sum of n is divisible by 5, the digit sum of n+1 can't physically be divisble by 5, unless we utilise 9's at the end, this way whenever we take a number in the tens (i.e. 19), the n+1 will be 1 off being divisble, so if we take a number in the hundreds, (109, remember it must have as many 9's at the end as possible) the n+1 will be 2 off being divisble, so continuing this into the thousands being three, tenthousands being 4, the hundred thousands will be 5 off (or also divisble by 5). So if we stick a 1 in the beginning (for the lowest value), and fill the last digits with 9's, we by process of elimination realise that the tenthousands digit must be 3 such that the digit sum is divisible by 5, therefore we get 139,999

6 0
2 years ago
Can someone help me, please ?? im stuck 
Tresset [83]

Part A

5x - 5 >= 10 or-3x + 1 > 13

1) This is a disjunction (OR)

5x - 5 >= 10

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Answer

x <- 4 or x >= 3


Part C:

1) This is a conjunction (And)

5x+ 3 <= 18 and 4 - x < 6

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Answer

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Answer:

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3 0
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