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tigry1 [53]
4 years ago
10

Which linear equations does the graph show the solution to? Select all that apply. y = 2x + 13 y = –x – 2 y = 3x – 5 y = negativ

e one-half x + 6 y = –2x – 2

Mathematics
2 answers:
polet [3.4K]4 years ago
8 0

Answer:

y = 2x + 13

y = –x – 2

Step-by-step explanation:

salantis [7]4 years ago
6 0

Answer:

y = 2x + 13

y = –x – 2

Step-by-step explanation:

The options are:

y = 2x + 13

y = -x - 2

y = 3x - 5

y= -(1/2)x + 6

y = -2x - 2

The graph is shown in the figure attached. There, point (-5, 3) is shown. Replacing it into the equations we get:

y = 2(-5) + 13  = 3    (so, it is a solution)

y = -(-5) - 2 = 3     (so, it is a solution)

y = 3(-5) - 5  = -20 ≠ 3   (so, it isn't a solution)

y= -(1/2)(-5) + 6  = 8.5 ≠ 3     (so, it isn't a solution)

y = -2(-5) - 2 = 8 ≠ 3     (so, it isn't a solution)

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Answer:

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Step-by-step explanation:

We are given that The daily high temperature X in degrees Celsius in Montreal during April has expected value E(X) = 10.3°C with a standard deviation SD(X) = 3.5°C.

The conversion of X into degrees Fahrenheit Y is Y = (9/5)X + 32.

(1) Y = (9/5)X + 32

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           = (9/5) * E(X) + 32   {\because expectation of constant is constant}

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Therefore, E(Y), the expected daily high in Montreal during April in degrees Fahrenheit is 50.54°F .

(2) Y = (9/5)X + 32

    SD(Y) = SD((9/5)X + 32) = SD((9/5)X) + SD(32)

              = (9/5)^{2} * SD(X) + 0  {\because standard deviation of constant is zero}

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Therefore, SD(Y), the standard deviation of the daily high temperature in Montreal during April in degrees Fahrenheit is 11.34°F .

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