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Kryger [21]
3 years ago
14

Find the slope intercept form of the liner function using the points (-6,6) and (-2,4) Show all work

Mathematics
1 answer:
olga2289 [7]3 years ago
4 0

y = - \frac{1}{2} x + 3

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y-intercept )

to calculate m use the gradient formula

m = ( y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (- 6, 6 ) and (x₂, y₂ ) = (- 2, 4 )

m = \frac{4-6}{-2+6} = \frac{-2}{4} = - \frac{1}{2}

y = - \frac{1}{2} x + c is the partial equation

to find c substitute either of the 2 points into the partial equation

using (- 2, 4 ), then

4 = 1 + c ⇒ c = 4 - 1 = 3

y = - \frac{1}{2} x + 3 ← in slope-intercept form


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Rewa Delta Union Rugby CEO has become concerned about the slow pace of the rugby games played in the current union rugby, fearin
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Answer:

We conclude that the mean duration of 15-sided union rugby games has decreased after the meeting.

Step-by-step explanation:

We are given that Before the meeting, the mean duration of the 15-sided rugby game time was 3 hours, 5 minutes, that is, 185 minutes.

A random sample of 36 of the 15-sided rugby games after the meeting showed a mean of 179 minutes with a standard deviation of 12 minutes.

Let \mu = <em><u>mean duration of 15-sided union rugby games after the meeting.</u></em>

So, Null Hypothesis, H_0 : \mu \geq 185 minutes      {means that the mean duration of 15-sided union rugby games has increased or remained same after the meeting}

Alternate Hypothesis, H_A : \mu < 185 minutes     {means that the mean duration of 15-sided union rugby games has decreased after the meeting}

The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about the population standard deviation;

                            T.S. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean duration of 15-sided union rugby games = 179 min

            s = sample standard deviation = 12 minutes

            n = sample of 15-sided rugby games = 36

So, <u><em>the test statistics</em></u>  =  \frac{179-185}{\frac{12}{\sqrt{36} } }  ~ t_3_5

                                       =  -3

The value of t test statistics is -3.

<u>Now, at 1% significance level the t table gives critical value of -2.437 at 35 degree of freedom for left-tailed test.</u>

Since our test statistic is less than the critical value of t as -3 < -2.437, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that the mean duration of 15-sided union rugby games has decreased after the meeting.

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Answer:

\frac{-5 \pm \sqrt{13} }{6}

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
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Step-by-step explanation:

<u>Step 1: Define</u>

<u />\frac{-5 \pm \sqrt{5^2-4(3)(1)} }{2(3)}<u />

<u />

<u>Step 2: Evaluate</u>

  1. Evaluate Exponents:                    \frac{-5 \pm \sqrt{25-4(3)(1)} }{2(3)}
  2. Evaluate Multiplication:               \frac{-5 \pm \sqrt{25-12} }{6}
  3. Evaluate Subtraction:                  \frac{-5 \pm \sqrt{13} }{6}
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