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frosja888 [35]
3 years ago
7

Use the given conditions to write an equations for the line in slope- intercept form. passing through (1,-8) and (-7,8)

Mathematics
1 answer:
Hoochie [10]3 years ago
6 0

Answer:

y = -2x - 6

Step-by-step explanation:

Going from the first point to the second, we see x decreasing by 8 from 1 to -7 (this is the 'run') and y increasing by 16 from -8 to +8 (this is the 'rise').  Thus, the slope of the line through these two points is m = rise/run =  16/(-8) = -2.

Using the point-slope formula y - k = m(x - h) and the point (1, -8), we get:

y + 8 = -2(x - 1), or

y = -8 - 2x + 2, or

y = -2x - 6 (in slope-intercept form)

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Step-by-step explanation:

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First step to solving the equation is

By multiplying both sides by the value of 5.

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A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime
Shalnov [3]

Answer:

We conclude that the true average percentage of organic matter in such soil is different from 3%.

Step-by-step explanation:

We are given that the values of the sample mean and sample standard deviation are 2.481 and 1.616, respectively.

Suppose we know the population distribution is normal, we have to test the hypothesis that does this data suggest that the true average percentage of organic matter in such soil is something other than 3%.

<em>Let </em>\mu<em> = true average percentage of organic matter in such soil</em>

SO, <u>Null Hypothesis</u>, H_0 : \mu = 3%   {means that the true average percentage of organic matter in such soil is equal to 3%}

<u>Alternate Hypothesis</u>, H_A : \mu \neq 3%   {means that the true average percentage of organic matter in such soil is different than 3%}

The test statistics that will be used here is <u>One-sample t test statistics</u> because 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 amount of organic matter = 2.481%

              s = sample standard deviation = 1.616%

              n = sample of soil specimens = 30

So, <u><em>test statistics</em></u>  =  \frac{0.02481-0.03}{{\frac{0.01616}{\sqrt{30} } } }  ~ t_2_9

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<u></u>

<u>Now, P-value of the test statistics is given by;</u>

       P-value = P( t_2_9 > -1.759) = <u>0.046</u> or 4.6%

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<em>Now, here the P-value is 0.046 which is clearly smaller than the level of significance of 0.05 (for two-tailed test), so we will reject our null hypothesis as it will fall in the rejection region.</em>

Therefore, we conclude that the true average percentage of organic matter in such soil is different from 3%.

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