3 years ago

Whats the partial products for 15x32

Mathematics

1 answer:

Bumek [7]3 years ago

8 0

Answer:

160 & 320

Step-by-step explanation

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Answer:

Step-by-step explanation:

3RD MAN: 1

2ND MAN: 2

1ST MAN: 4

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Step-by-step explanation:

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4 years ago

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Consider rolling two fair dice and observing the number of spots on the resulting upward face of each one. Letting A be the even

Allushta [10]

Answer:

$P(E|A)= \frac{10}{11}$

Step-by-step explanation:

Given

Two rolls of die

$E \to$ one of the outcomes is 6

$A \to$ atleast one is 6

Required

P(E|A)

First, list out the outcome of each

$E = \{(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5)\}$

$A = \{(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$

So:

$P(E|A)= \frac{n(E\ n\ A)}{n(A)}$

Where:

$E\ n\ A = \{(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5)\}$

$n(E\ n\ A) = 10$

$n(A) = 11$

So:

$P(E|A)= \frac{10}{11}$

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