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Olegator [25]
3 years ago
7

Find the slope-intercept form of an equation of the line perpendicular to the graph of x – 3y = 5 and passing through (0, 6). Qu

estion 3 options: a) y = –3x + 6 b) y = 13x − 2 c) y = 13x + 2 d) y = 3x – 6
Mathematics
2 answers:
patriot [66]3 years ago
7 0

Hello there,

Well first we need to write down what we know/what are we finding out...

 - the original line has the equation of x - 3y = 5

 - we are finding the equation of the line that is perpendicular

Now we start by changing the original equation to slope-intercept form:

  • x - 3y = 5 as slope-intercept form would be y = 1/3x -5/3

Perpendicular lines have the negative reciprocal slopes so the new line has to have the slope as -3.

Option A would be the correct answer: a) y = -3x + 6

Hope I helped,

Amna

Nikitich [7]3 years ago
3 0

Answer:

a) y=-3x+6

Step-by-step explanation:

finding the slope of the given line:

-3y=-x+5

y=(1/3)x-(5/3)

since our new line has to be perpendicular to this line, our new line will have slope m=-3

putting these quantities into point-slope form:

y-6=-3(x-0)

converting to slope-int:

y=-3x+6

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snow_tiger [21]

Answer:

2 < x < 3

Step-by-step explanation:

4x - 4 < 8    AND    9x + 5 > 23

4x < 12  AND   9x > 18

x < 3 AND x > 2 so the answer is 2 < x < 3.

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What are the solutions of -1 ≤ x – 3 &lt; 4
cricket20 [7]

Answer:

2 ≤ x  < 7

Step-by-step explanation:

-1 ≤ x – 3 < 4

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3 years ago
A coffee packaging plant claims that the mean weight of coffee in its containers is at least 32 ounces. A random sample of 15 co
Luda [366]

Answer:

We conclude that the mean weight of coffee in its containers is at least 32 ounces which means that the data support the claim.

Step-by-step explanation:

We are given that a coffee packaging plant claims that the mean weight of coffee in its containers is at least 32 ounces.

A random sample of 15 containers were weighed and the mean weight was 31.8 ounces with a sample standard deviation of 0.48 ounces.

Let \mu = <u><em>mean weight of coffee in its containers.</em></u>

SO, Null Hypothesis, H_0 : \mu \geq 32 ounces     {means that the mean weight of coffee in its containers is at least 32 ounces}

Alternate Hypothesis, H_A : \mu < 32 ounces      {means that the mean weight of coffee in its containers is less than 32 ounces}

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

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

where, \bar X = sample mean weight = 31.8 ounces

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             n = sample of containers = 15

So, <u><em>the test statistics</em></u>  =  \frac{31.8 -32}{\frac{0.48}{\sqrt{15} } }  ~ t_1_4  

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The value of t test statistics is -1.614.

<u>Now, at 0.01 significance level the t table gives critical value of -2.624 at 14 degree of freedom for left-tailed test.</u>

Since our test statistic is more than the critical value of t as -1.614 > -2.624, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u><em>we fail to reject our null hypothesis</em></u>.

Therefore, we conclude that the mean weight of coffee in its containers is at least 32 ounces which means that the data support the claim.

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