All categories

3 years ago

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

You might be interested in

30 points!!!! Please help!!!!

lara31 [8.8K]

Answer:

T = 2π√[2m/(9.8m/s²)]

5 0

3 years ago

An employee makes 9/hour. The employee arrives ten minutes late each day. How much money does the employee lose each week?

Dmitriy789 [7]

The employee makes 7.50 per hour

7 0

3 years ago

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

ozzi

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 $\mu$ = mean duration of 15-sided union rugby games after the meeting.

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 One-sample t test statistics 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, the test statistics = $\frac{179-185}{\frac{12}{\sqrt{36} } }$ ~ $t_3_5$

= -3

The value of t test statistics is -3.

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

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 we reject our null hypothesis.

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

6 0

3 years ago

PLEASE HELP ASAP Solve. y > 2x + 2 y ≤ -3/2 x - 5

Fynjy0 [20]

$2x + 2 < - \frac{3}{2} x - 5 \\ 2x + \frac{3}{2} x < - 5 - 2 \\ 3.5x < - 7 \\ x< - 2$
$x < - 2 \\ x < \frac{y - 2}{2}$
$-2 = \frac{y - 2}{2} \\ y=-2$