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Zepler [3.9K]
3 years ago
9

Help me with this please.

Mathematics
1 answer:
jek_recluse [69]3 years ago
3 0
I don’t know answer but I use gauthmath to help me with math u may as well use it and u can use this code P9EJ5H
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4 years ago
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Consider rolling two fair dice one 3-sided the other 5-sided
Ne4ueva [31]

Since the dice are fair and the rolling are independent, each single outcome has probability 1/15. Every time we choose

1\leq x\leq 3,\quad 1\leq y \leq 5

We have P(X=x)=\frac{1}{3} and P(Y=y)=\frac{1}{5}, because the dice are fair.

Now we use the assumption of independence to claim that

P(X=x, Y=y) = P(X=x)\cdot P(Y=y) =\dfrac{1}{3}\cdot\dfrac{1}{5} = \dfrac{1}{15}

Now, we simply have to count in how many ways we can obtain every possible outcome for the sum. Consider the attached table: we can see that we can obtain:

  • 2 in a unique way (1+1)
  • 3 in two possible ways (1+2, 2+1)
  • 4 in three possible ways
  • 5 in three possible ways
  • 6 in three possible ways
  • 7 in two possible ways
  • 8 in a unique way

This implies that the probabilities of the outcomes of W=X+Y are the number of possible ways divided by 15: we can obtain 2 and 8 with probability 1/15, 3 and 7 with probability 2/15, and 4, 5 and 6 with probabilities 3/15=1/5

8 0
3 years ago
What is the answer? Please explain.<br><br> (5x^2 -3) + (2x^2 - 3x^3)
ale4655 [162]

Answer:

(5x^2-3) +(2x^2-3x^3)

=3x^3+5x^2+2x^2-3

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Step-by-step explanation:

hope this is correct cuz I'm not an expert...

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3 years ago
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