Asymptote at x = 3 and a horizontal asymptote at y = 1. The curves approach these asymptotes but never cross them. The method used to find the horizontal asymptote changes depending on how the degrees of the polynomials in the numerator and denominator of the function compare.
Answer:
x^5
Step-by-step explanation:
<h3>
Answer: No, this function is not linear</h3>
This function is a hyperbola. It graphs out two disjoint curves. A linear function produces a single straight line graph. All linear functions can be written in the form y = mx+b. The x being in the denominator is one indicator we cannot write the original equation in the form y = mx+b.
y = 3/x is the same as xy = 3; this tells us every point on y = 3/x has its (x,y) coordinate pair multiply to 3.
Answer:
The IQR is given by:
If we want to find any possible outliers we can use the following formulas for the limits:
And if we find the lower limt we got:
So then the left boundary for this case would be 3 days
Step-by-step explanation:
For this case we have the following 5 number summary from the data of 144 values:
Minimum: 9 days
Q1: 18 days
Median: 21 days
Q3: 28 days
Maximum: 56 days
The IQR is given by:
If we want to find any possible outliers we can use the following formulas for the limits:
And if we find the lower limt we got:
So then the left boundary for this case would be 3 days
75.
Explanation: from 3 to 12, you add 9, and then double the 12 to get to 24, and then add 9 again to get to 33, and double it to get 66, so the pattern is to add 9 and then double.
