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Oksana_A [137]
2 years ago
8

What is the answer?

Mathematics
1 answer:
ValentinkaMS [17]2 years ago
8 0

Answer:

D

Step-by-step explanation:

slope =4, y intercept = -6

sorry I currently don't have enough time to explain the calculation

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The graph of the piecewise function f(x) is shown what is the domain of f(x)
Rom4ik [11]

The domain of a function is where we have a value for x.

Since that's the case the domain of f(x) = {x e R / 1 ≤ x < 5}

We see that we have a value for x = 1 cuz we have a filled circle, but we don't have a value for x = 5, look at the unfilled circle

So, our x can vary between 1 and 5, but can't be 5.

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Solve for t <br> 1/2t + 6 = -7
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1/2t+6=-7
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Let f(x) = x2 − 16. find f−1(x). open study
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Okay, so first you make f(x) = y and switch x and y. So the equation becomes:

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the square root of x + 4 = y
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What is the value of x in the inequality 7 minus 2x/ -4 + 2 &lt; -x?
Fantom [35]
Maybe it was (7-2x)/-4+2<-x then times 4 both sides <span> </span> -7+2x+8<-4x <span> </span> 2x+1<-4x <span> </span> 6x+1<0 <span> </span> 6x<-1 <span> </span> x<-1/6 <span> </span> there we go
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A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday�s mail. In actuali
Softa [21]

Answer:

P(Y = 0) = 0.09

P(Y = 1) = 0.4

P(Y = 2) = 0.32

P(Y = 3) = 0.19

Step-by-step explanation:

Let the events be:

W = Wednesday

T = Thursday

F = Friday

S = Saturday

Their corresponding probabilities are

P(W) = 0.3\\P(T) = 0.4\\P(F) = 0.2\\P(S) = 0.1

Since Y = number of days beyond Wednesday that it takes for both magazines to arrive(so possible Y values are 0, 1, 2 or 3)

The possible number of outcomes are therefore 4^2 = 16\\(W, W), (W, T), (W, F), (W, S)\\(T, W), (T, T), (T, F), (T, S)\\(F, W), (F, T), (F, F), (F, S)\\(S, W), (S, T), (S, F), (S, S)

The values associated for each of the outcomes are as follows:

Y(W, W) = 0, Y(W, T) = 1, Y(W, F) = 2, Y(W, S) = 3\\Y(T, W) = 1, Y(T, T) = 1, Y(T, F) = 2, Y(T, S) = 3\\Y(F, W) = 2, Y(F, T) = 2, Y(F, F) = 2, Y(F, S) = 3\\Y(S, W) = 3, Y(S, T) = 3, Y(S, F) = 3, Y(S, S) = 3

The probability mass function of Y is,

P(Y = 0) = 0.3(0.3) = 0.09\\P(Y = 1) = P[(W, T) or (T, W) or (T, T)]\\= [0.3(0.4) + 0.3(0.4) + 0.4(0.4)]\\= 0.4\\\\P(Y = 2) = P[(W, F) or (T, F) or (F, W) or (F, T) or (F, F)]\\= [0.3(0.2) + 0.4(0.2) + 0.2(0.3) + 0.2(0.4) + 0.2(0.2)]\\= 0.32\\\\P(Y = 3) = P[(W, S) or (T, S) or (F, S) or (S, W) or (S, T) or (S, F) or (S, S)]\\= [0.3(0.1) + 0.4(0.1) + 0.2(0.1) + 0.1(0.3) 0.1(0.4) + 0.1(0.2) + 0.1(0.1)]\\= 0.19

7 0
3 years ago
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