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Neko [114]
2 years ago
11

A cooler contains 5 cans of soda, 2 colas 2 orange and 1 cherry. Two cans are selected at random without replacement. Find the p

rob that at least one can is a cola.
A) 1/10
B) 3/10
C) 7/10
D) 9/10
E) None
Mathematics
1 answer:
konstantin123 [22]2 years ago
8 0

Answer:

9/10

Step-by-step explanation:

Probability of 1st can being cola = 2/5

If not a cola then probability of 2nd can being cola = 2/4 = 1/2

So total = 2/5 + 1/2 = 9/10.

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(x-2)^2-64=0\\(x-2)^2=64\\x-2=8 \vee x-2=-8\\x=10 \vee x=-6

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wolverine [178]

Using the determinant method, the cross product is

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Or you can apply the properties of the cross product. By distributivity, we have

(3i + 8j - 6k) x (-4i - 2j - 3k)

= -12(i x i) - 32(j x i) + 24(k x i) - 6(i x j) - 16(j x j) + 12(k x j) - 9(i x k) - 24(j x k) + 18(k x k)

Now recall that

  • (i x i) = (j x j) = (k x k) = 0 (the zero vector)
  • (i x j) = k
  • (j x k) = i
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Putting these rules together, we get

(3i + 8j - 6k) x (-4i - 2j - 3k)

= -32(-k) + 24j - 6k + 12(-i) - 9(-j) - 24i

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D . Linear

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