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goblinko [34]
2 years ago
15

Write the standard form of the line that passes through the given points. include your work in your final answer. (-1, -3) and (

2, 1)
Mathematics
1 answer:
yan [13]2 years ago
6 0

The standard form of a line passing through the points (-1, -3) and (2, 1) is <u>4x - 3y = 5</u>.

The slope of the given line, m = (1 - (-3))/(2 - (-1)) = (1 + 3)/(2 + 1) = 4/3.

Computed using the formula for the slope of a line, m = (y₂ - y₁)/(x₂ - x₁), when a line passes through the points (x₁, y₁) and (x₂, y₂).

The point intercept form of a line is <u>y - y₁ = m(x - x₁)</u> when the line passes through the point (x₁, y₁) and has the slope m.

Thus, the given line in the point intercept form can be written as:

y - 1 = (4/3)(x - 2).

The standard form of a line is ax + by = c.

To convert the point intercept form to the standard form, we do as follows:

y - 1 = (4/3)(x - 2),

or, 3(y - 1) = 3(4/3)(x - 2) {Multiplying both sides by 3},

or, 3y - 3 = 4x - 8 {Simplifying},

or, 8 - 3 = 4x - 3y {Rearranging},

or, 4x - 3y = 5 {Rearranging and simplifying}.

Thus, the standard form of a line passing through the points (-1, -3) and (2, 1) is <u>4x - 3y = 5</u>.

Learn more about equations of straight lines at

brainly.com/question/13763238

#SPJ4

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Answer:

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R output: Call:

lm(formula = y ~ x1 + x2)

Residuals:  

   Min      1Q Median      3Q     Max

-41.730 -12.174   0.791 12.374 40.093

Coefficients:

        Estimate Std. Error t value Pr(>|t|)    

(Intercept) 415.113     82.517   5.031 0.000709 ***  

x1            -6.593      4.859 -1.357 0.207913    

x2            -4.504      1.071 -4.204 0.002292 **  

---  

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1  

Residual standard error: 24.45 on 9 degrees of freedom  

Multiple R-squared: 0.768,     Adjusted R-squared: 0.7164  

F-statistic: 14.9 on 2 and 9 DF, p-value: 0.001395

a).  y=415.113 +(-6.593)x1 +(-4.504)x2

b). s=24.45

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P=0.0014

There is convincing evidence at least one of the explanatory variables is significant predictor of the response.

e).  newdata=data.frame(x1=21.3, x2=43)

# confidence interval

predict(mod, newdata, interval="confidence")

#prediction interval

predict(mod, newdata, interval="predict")

confidence interval

> predict(mod, newdata, interval="confidence",level=.95)

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1 81.03364 43.52379 118.5435

95% CI = (43.52, 118.54)

f).  #prediction interval

> predict(mod, newdata, interval="predict",level=.95)

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