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juin [17]
3 years ago
9

Which equation represents the line that passes through the points (-3, 7) and (9,-1)? oy--3x+5 o y=-x-7 o y=zx-7 oy - 12 2x+5​

Mathematics
1 answer:
nekit [7.7K]3 years ago
4 0

Answer:

use the slope equation (y2-y1) / (x2 -x1). plug in values and solve

(-1 -7) / (9 -(-3)) (subtracting a negative is the same as adding a positive)

-8 / 12 (simplify)

m = -2/3

then, plug in one of the points and the slope into the slope-point equation.

(y - y1) = m (x - x1) (the point-slope form)

y - 7 = -2/3( x - (-3)) (plug in the slope and point 1 values, then solve)

y - 7 = -2/3x - 2 (add 7 to both sides)

y = 2/3x +5 (the answer)

Step-by-step explanation:

I hope this helps :))

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A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past
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Answer:

We conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.

Step-by-step explanation:

We are given that a particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8000.

From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9800 miles.

<u><em>Let </em></u>\mu<u><em> = average miles for deluxe tires</em></u>

So, Null Hypothesis, H_0 : \mu \geq 50,000 miles   {means that deluxe tire averages at least 50,000 miles before it needs to be replaced}

Alternate Hypothesis, H_A : \mu < 50,000 miles    {means that deluxe tire averages less than 50,000 miles before it needs to be replaced}

The test statistics that will be used here is <u>One-sample z test statistics</u> as we know about population standard deviation;

                                  T.S.  = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean lifespan = 46,500 miles

            \sigma = population standard deviation = 8000 miles

            n = sample of tires = 28

So, <u><em>test statistics</em></u>  =  \frac{46,500-50,000}{\frac{8000}{\sqrt{28} } }

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The value of the test statistics is -2.315.

Now at 5% significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is less than the critical value of z as -2.315 < -1.6449, 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 deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.

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