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Ilia_Sergeevich [38]
3 years ago
10

What is the range of the function represented by these ordered pairs? {(–2, 1), (0, 0), (3, –1), (–1, 7), (5, 7)}

Mathematics
1 answer:
qaws [65]3 years ago
5 0
The range is the set of y coordinates (or outputs). For any point (x,y) the y coordinate is the second number listed. For example, with (8,7) the y coordinate is 7.

So we simply list all of the y coordinates of each ordered pair. Those values are: 1, 0, -1, 7, 7

Let's order them as you would see on a number line

-1, 0, 1, 7, 7

then let's toss out the extra 7 to get

-1, 0, 1, 7

So the range is the set {-1, 0, 1, 7}
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U and V are mutually exclusive events. P(U) = 0.27; P(V) = 0.56. Find:
Alisiya [41]

Answer:

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

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

(c) P (U or V) = 0.83.

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.

7 0
3 years ago
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gulaghasi [49]
<span>Which numbers are divisible by nine?
107 513 944 126 802

answer is 513
because 513/9 = 57</span>
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Step-by-step explanation:

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