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4 years ago

What is the equation for this line?

Mathematics

1 answer:

zzz [600]4 years ago

7 0

Answer:

The equation of the line is y = 2x - 3

Step-by-step explanation:

The form of the linear equation is y = m x + b, where

m is the slope of the line, $m=\frac{y2-y1}{x2-x1}$

b is the y-intercept (value y at x = 0)

Let us take two points on the line to find its equation

∵ Points (3, 3) and (0, -3) lie on the line

∴ x1 = 3 and y1 = 3

∴ x2 = 0 and y2 = -3

→ Use the rule of the slope above to find m

∵ $m=\frac{-3-3}{0-3}=\frac{-6}{-3}=2$

∴ m = 2

→ Substitute its value in the form of the equation above

∴ y = 2x + b

∵ b is the value of y at x = 0

∵ y = -3 at x = 0

∴ b = -3

→ Substitute it in the equation above

∴ y = 2x + -3

→ Remember (+)(-) = (-)

∴ y = 2x - 3

∴ The equation of the line is y = 2x - 3

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$P(U |D ) = 0.198$

$P(O\ n \ B) = 0.188$

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The total number of deaths is mathematically represented as

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The total number of deaths in 1996 - 2000 is mathematically represented as

$T_a = 16 + 23+ \cdots + 30$

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The total number of deaths in 2001 - 2005 is mathematically represented as

$T_b = 17 + 16 + \cdots + 23$

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The total number of deaths in 2006 - 2010 is mathematically represented as

$T_c = 15 + 17 + \cdots + 16$

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Generally the the probability that it would occur under the tree given that the death was after 2000 is mathematically represented as

$P(U |D ) = \frac{P(A \ n\ U )}{P(A)}$

Here $P(A \ n\ U )$ represents the probability that it was after 2000 and it was under the tree and this is mathematically represented as

$P(A \ n\ U ) = \frac{Z}{ T}$

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$Z = 35 + 42$

=> $Z = 77$

=> $P(A \ n\ U ) = \frac{77}{ 623}$

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$P(A)$ is the probability of the death occurring after 2000 and this is mathematically represented as

$P(A) = \frac{T_b + T_c}{ T}$

=> $P(A) = \frac{ 206+ 182}{623}$

$P(U |D ) = \frac{\frac{77}{ 623} }{ \frac{ 206+ 182}{623}}$

=> $P(U |D ) = 0.198$

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$P(O | B) = \frac{T_z}{ T}$

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$P(O \ n \ B) = \frac{117}{623}$

=> $P(O\ n \ B) = 0.188$

Generally the probability that the death was from camping or being outside given that it was before 2001 is mathematically represented as

$P(O | B) = \frac{ P( O \ n \ B)}{ P(B)}$

Here $P(B)$ is the probability that it was before 2001 , this is mathematically represented as

$P(B ) = \frac{T_a}{T}$

=> $P(B ) = \frac{235}{623}$

$P(O | B) = \frac{ \frac{117}{623}}{ \frac{235}{623}}$

=> $P(O | B) = 0.498$

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