Which graph represents a line with a slope of and a y-intercept equal to that of the line y = 2/3 x – 2?
2 answers:
The graph should resemble the one I attached here. If the image is not visible, the line passes through the y-intercept at -2 and the x-intercept at 3.
Explanation: The y-intercept is clearly -2. Using the slope expressed as rise over run, the line rises 2 units and runs 3 units to the right. You can also reverse this by going down 2 units and left 3 units. Repeating these steps twice in both directions will trace the entire line.
Explanation: The y-intercept is clearly -2. Using the slope expressed as rise over run, the line rises 2 units and runs 3 units to the right. You can also reverse this by going down 2 units and left 3 units. Repeating these steps twice in both directions will trace the entire line.
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Response:
Detailed explanation:
Greetings!
You have the variable
X: Area eligible for painting with a can of spray paint (feet²)
This variable is normally distributed with a mean of μ= 25 feet² and a standard deviation of δ= 3 feet²
As this variable has a normal distribution, it needs to be converted into the standard normal form to utilize tabulated cumulative probabilities.
a.
P(X>27)
The first step involves standardizing the X value using Z= (X-μ)/ δ ~N(0;1)
P(Z>(27-25)/3)
P(Z>0.67)
Having determined the Z value, you can find it in the table, but since the table includes probabilities for , the following conversion must be applied:
P(Z>0.67)= 1 - P(Z≤0.67)= 1 - 0.74857= 0.25143
b.
A sample of 20 cans was taken, and you need to ascertain the probability of averaging a coverage area of 540 feet².
The sample mean maintains the same distribution as its source variable, but its variance is influenced by sample size, thus it is normally distributed with parameters:
X[bar]~N(μ;δ²/n)
To cover 540 feet² with 20 cans, the average coverage must be approximately 540/20= 27 feet² per can.
c.
P(X≤27) = P(Z≤(27-25)/(3/√20))= P(Z≤2.98)= 0.999d.No, if the distribution is not normal and skewed, the normal distribution should not be applied for calculating probabilities. While the central limit theorem might approximate the sampling distribution to normal when the sample size is 30 or larger, that isn’t applicable here.
I trust this information is helpful!
Response:
the solution can be found in the image
