Mathematics College

Does anyone know what the slope equation is for this graph. PLz explain if you can. Ty!

Answers

Answer 1

Answer: y = (-3/2)x+3

In decimal form, this is y = -1.5x+3

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Explanation:

Start at (0,3). Move three units down and two units to the right to get to (2,0)

slope = rise/run

slope = -3/2

rise = -3 = go 3 units down

run = 2 = go 2 units to the right

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We could use the two points (0,3) and (2,0) in the slope formula below

m = (y2-y1)/(x2-x1)

m = (0-3)/(2-0)

m = -3/2

So this is another way to see the slope is -3/2

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The y intercept is where the graph crosses the vertical y axis. In this case, it's at 3. So b = 3

Using m = -3/2 and b = 3, we go from y = mx+b to y = (-3/2)x+3 which is equivalent to y = -1.5x+3

Answer 2

Answer:

[tex]y=-\frac{3}{2} x+3[/tex]

Step-by-step explanation:

I don't really know how to explain this, but I hope the graph helps some.

I hope this helps you! (✿◕‿◕✿)

Related Questions

A food-protection agency counts the number of insect heads found per 100-gram batch of wheat flour. The researchers have 500 batches, and they want to know whether the frequency of insect heads in batches follows a distribution called a Poisson distribution. To generate the expected frequencies of batches with different numbers of insect heads under a Poisson distribution, they had to estimate the mean number of insect heads per batch from the data. The 500 batches included at least 5 batches having 0, 1, 2, 3, or 4 insect heads. No batches had more than four heads. Given this information, there are_____classes (k).

Answers

Answer:

The correct answer is k = 5.

Step-by-step explanation:

The Chi-square goodness of fit test would be used to determine whether the frequency of insect heads in batches follows a distribution called a Poisson distribution.

The hypothesis for the test can be defined as follows:

H₀: The frequency of insect heads in batches does not follows a Poisson distribution.

Hₐ: The frequency of insect heads in batches follows a Poisson distribution.

Assume that the significance level of the test is, α = 0.05.

The Chi-square test statistic is given by:

[tex]\chi^{2}=\sum\limits^{k}_{i=1}\frac{(O_{i}-E_{i})^{2}}{E_{i}}[/tex]

It is provided that the 500 batches included at least 5 batches having 0, 1, 2, 3, or 4 insect heads. No batches had more than four heads.

So, the number of classes or categories are, k = 5.

Thus, the correct answer is k = 5.

A process is normally distributed and in control, with known mean and variance, and the usual three-sigma limits are used on the control chart, so that the probability of a single point plotting outside the control limits when the process is in control is 0.0027. Suppose that this chart is being used in phase I and the averages from a set of m samples or subgroups from this process are plotted on this chart. What is the probability that at least one of the averages will plot outside the control limits when m = 5? Repeat these calculations for the cases where m = 10, m = 20, m = 30, and m = 50.

Required:
Discuss the results that you have obtained.

Answers

Answer:

The solution to the issue is outlined in the following portion of the summary.

Step-by-step explanation:

The given value is:

p = 0.0027

When m = 5,