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3 years ago

What is the probability of rolling a sum of 5 if two regular 6-sided number cubes are rolled?

Mathematics

1 answer:

creativ13 [48]3 years ago

7 0

1/9 because 5 shows up 4 times out of 36 times so that would be 4/36 and you would simplify it to get 1/9

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Answer:

slope: 3 for N and 1/2 for O

Step-by-step explanation:

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3 years ago

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13. Given that

AlladinOne [14]

step by step explanation:

$\mathfrak{x}^{2}+{y}^{2}+16=0$

=[x2+16=0x26]

=[2x{y}^2{16}~0]

=[4×{y}^0{16}]

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2 years ago

Caleb started saving money in a cookie jar. He started with $25 .He adds $10 to the cookie jar each week. Write an equation wher

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3 years ago

You have 16 coins that are heads up and 18 coins that are tails up. After you add some coins that are tails up, the ratio of hea

Gwar [14]

So 16=heads
18=tails
after you add some tails
18+x

ratio of heads to tails is 1:1.5
heads=16, didn't change
tails=18+x

so
1:1.5=16:18+x
fractionify
1/1.5=16/(18+x)
1/1.5=2/3
2/3=16/(18+x)
mutiply both sides by (3)(18+x)
2(18+x)=3(16)
36+2x=48
subtract 36 from both sides
2x=12
divide 2
x=6

6 tails added

total=heads+tails
heads=16
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total=16+24
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6 tails added
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2 years ago

Suppose I am playing a sport that does not allow me to end a match in a tie (AKA I have to either win or lose). If my probabil

Leviafan [203]

Answer: The required probability is 0.3456.

Step-by-step explanation:

Since we have given that

Probability of winning at any given time = 0.6

Probability of losing at any given time = 1-0.6 = 0.4

Number of total matches = 5

Number of won matches = 3

So, using "Binomial distribution", we get that

$P(X=3)=^5C_3(0.6)^3(0.4)^2\\\\P(X=3)=0.3456$

Hence, the required probability is 0.3456.

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3 years ago