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iren [92.7K]
4 years ago
9

What is the slope of the line described by the equation below? Y - 5= -3(x - 17)

Mathematics
2 answers:
MA_775_DIABLO [31]4 years ago
7 0
Hello there!

To find the slope, we must first put this equation in y = mx + b form (also known as slope intercept form)

First, let's solve the right side using the distributive property.

To do this, we will multiply -3 by both x and -17

-3 * x = -3x

-3 * -17 = 51 (a negative times a negative equals a positive)

So, now we have: y - 5 = -3x + 51


Now we must get rid of the -5 on the right side.

To do this, we will add -5 to both sides

y - 5 + 5 = -3x + 51 + 5

Now we have y = -3x + 56

Now, we know the slope is -3

I hope my answer helped you out!

~ Fire
Alenkinab [10]4 years ago
4 0
First distribute -3 to both x and -17
-3(x) = -3x
-3(-17) = 51

y - 5 = -3x + 51

isolate the y, add 5 to both sides

y - 5 (+5) = -3x + 51 (+5)

add all numbers with same variables together

y = -3x + 51 + 5
y = -3x + 56

inside the slope intercept form:

y = y
m = slope
x = x
b = intercept.

y = mx + b (using this equation, we can see that -3 = m)

-3 is slope, and is your answer 

hope this helps
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Suppose X, Y, and Z are random variables with the joint density function f(x, y, z) = Ce−(0.5x + 0.2y + 0.1z) if x ≥ 0, y ≥ 0, z
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The value of the constant C is 0.01 .

Step-by-step explanation:

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Suppose X, Y, and Z are random variables with the joint density function,

f(x,y,z) = \left \{ {{Ce^{-(0.5x + 0.2y + 0.1z)}; x,y,z\geq0  } \atop {0}; Otherwise} \right.

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\int_x( {\int_y( {\int_z {f(x,y,z)} \, dz }) \, dy }) \, dx = 1

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C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y}[0+\frac{1}{0.1}]  } \, dy  }) \, dx =1

10C\int\limits^\infty_0 {e^{-0.5x}([\frac{-e^{-0.2y} }{0.2}]^\infty__0  }) \, dx = 1

10C\int\limits^\infty_0 {e^{-0.5x}([\frac{-e^{-0.2(\infty)} }{0.2}+\frac{e^{-0.2(0)} }{0.2}]   } \, dx = 1

10C\int\limits^\infty_0 {e^{-0.5x}[0+\frac{1}{0.2}]  } \, dx = 1

50C([\frac{-e^{-0.5x} }{0.5}]^\infty__0}) = 1

50C[\frac{-e^{-0.5(\infty)} }{0.5} + \frac{-0.5(0)}{0.5}] =1

50C[0+\frac{1}{0.5} ] =1

100C = 1 ⇒ C = \frac{1}{100}

C = 0.01

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