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Oliga [24]
3 years ago
6

Which is a perfect square a. 6^1 b. 6^2 c. 6^3 d. 6^5 (need answer now pls)

Mathematics
1 answer:
Bas_tet [7]3 years ago
6 0
Hello there! 6^1 is not a perfect square. It's basically it's own number and 6 is not a perfect square. Therefore, A is out. 6^3 (216) is cubed, not squared, so C is out as well. Just by looking at the exponents, 6^2 is squared and 36 is a perfect square, so the answer is B: 6^2.
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U and V are mutually exclusive events. P(U) = 0.27; P(V) = 0.56. Find:
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Answer:

(a) P (U and V) = 0.

(b) P (U|V) = 0.

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Step-by-step explanation:

Mutually exclusive events are those events that cannot occur at the same time. That is, if events A and B are mutually exclusive then,

                                         P (A\cap B) = 0

<u>Given</u>:

Events U and V are mutually exclusive.

P (U) = 0.27 and P (V) = 0.56

(a)

As events U and V are mutually exclusive, the probability of their intersection will be 0.

That is,

P(U\ and\ V) = P (U\cap V) = 0

Thus, the value of P (U and V) is 0.

(b)

The conditional probability of event B given A is:

P(B|A) =\frac{P(A\cap B)}{P(B)}

Compute the value of P (U|V) as follows:

P(U|V) =\frac{P(U\cap V)}{P(V)}\\=\frac{0}{0.56}\\ =0

Thus, the value of P (U|V) is 0.

(c)

The probability of the union of two events, say A and B, is

P(A\ or\ B)=P(A\cup B)=P(A)+P(B)-P(A\cap B)

Compute the value of P (U or V) as follows:

P(U\ or\ V)=P(U\cup V)\\=P(U)+P(V)-P(U\cap V)\\=0.27+0.56-0\\=0.83

Thus, the value of P (U or V) is 0.83.

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\bold{ANSWER:}
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