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3 years ago

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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Answer:

Step-by-step explanation:

Given the equation:

1/5 × (x - 2) = 20

First step to solving the equation is

By multiplying both sides by the value of 5.

That will give,

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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%.

Let $\mu$ = true average percentage of organic matter in such soil

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

Alternate Hypothesis, $H_A$ : $\mu \neq$ 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 One-sample t test statistics 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%

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P-value = P( $t_2_9$ > -1.759) = 0.046 or 4.6%

If the P-value of test statistics is more than the level of significance, then we will not reject our null hypothesis as it will not fall in the rejection region.

If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis as it will fall in the rejection region.

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.

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

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