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slavikrds [6]
4 years ago
6

Match each given domain of the function with its corresponding range. {40, 45, 50, 55, 60} {52, 56, 63, 67, 71} {37, 46, 49, 55,

57} {44, 51, 59, 66, 73} Range Domain {20.8, 22.4, 25.2, 26.8, 28.4} arrowBoth {16, 18, 20, 22, 24} arrowBoth {17.6, 20.4, 23.6, 26.4, 29.2} arrowBoth {14.8, 18.4, 19.6, 22, 22.8} arrowBoth
ppppllleeeaaassseee someone answer this its important.
Mathematics
1 answer:
inn [45]4 years ago
8 0

Answer:

RANGE: {20.8, 22.4, 25.2, 26.8, 28.4} goes with DOMAIN: {52, 56, 63, 67, 71}

RANGE: {16, 18, 20, 22,24} goes with DOMAIN: {40, 45, 50, 55, 60}

RANGE: {17.6, 20.4, 23.6, 26.4, 29.2} goes with DOMAIN: {44, 51, 59, 66, 73}

RANGE: {14.8, 18.4, 19.6, 22, 22.8} goes with DOMAIN: {37, 46, 49, 55, 57}

Step-by-step explanation:


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Answer:

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Step-by-step explanation:

x=8

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5 0
3 years ago
A large cheese pizza costs $18. Each topping you add on costs $1.50. How much would it cost to get a large cheese pizza with cc
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Answer:

\$18+\$1.50c

Step-by-step explanation:

Let

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7 0
4 years ago
Solve this equation for x. (x+4) (x-2)=0
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7 0
3 years ago
The failure of a circuit board interrupts work that utilizes a computing system until a new board is delivered. The delivery tim
Hatshy [7]

Answer:

The expected cost is 152

Step-by-step explanation:

Recall that since Y is uniformly distributed over the interval [1,5] we have the following probability density function for Y

f_Y(y ) = \frac{1}{5-1} = \frac{1}{4} if 1\leq y \leq 5 and 0 othewise. (To check this is the pdf, check the definition of an uniform random variable)

Recall that, by definition  

E(Y^k) = \int_{-\infty}^{\infty} y^kf_Y(y)dy

Also, we are given that C = 50+3Y+9Y^2. Recall the following properties of the expected value. If X,Y are random variables, then

E(a+bX+cY) = a+bE(X)+cE(Y)

Then, using this property we have that E[C] = 50+3E[Y]+ 9E[Y^2].

Thus, we must calculate E[Y] and E[Y^2].

Using the definition, we get that

E[Y] = \int_{1}^{5}\frac{y}{4} dy =\frac{1}{4}\left\frac{y^2}{2}\right|_{1}^{5} = \frac{25}{8}-\frac{1}{8} = 3

E[Y^2] = \int_{1}^{5}\frac{y^2}{4} dy =\frac{1}{4}\left\frac{y^3}{3}\right|_{1}^{5} = \frac{125}{12}-\frac{1}{12} = \frac{31}{3}

Then

E(C) = 50 + 3\cdot 3 + 9 \cdot \frac{31}{3} = 152

5 0
3 years ago
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Gnom [1K]

Answer:

(-5,4)

Step-by-step explanation:

8 0
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