The very first thing to do in every correlation activity is to plot the gathered data points in a scatter plot. It is better to use software tools like MS Excel because they have a feature there that uses linear regression like that one shown in the picture.
Once you plot the data points, make a trendline. You are given with options. If you want a linear function, then you will have a linear model with a function equation of y = 0.2907x + 2.2643. It has a correlation coefficient of 0.9595. That's a strong correlation already. The R² value tells how good your model fits the data points. If you want to increase the R², a better model would be a quadratic function with the equation, y = -0.0209x²+0.506x+2.0232. As you can see the R² increase even more to 0.9992.
Answer:
$45.12
Step-by-step explanation:
($0.47 /stamp)×(8 stamps/mo)×(12 mo/yr) = $45.12 /yr
You will save $45.12 per year on stamps.
Any number times 0 is going to equal 0
We use the trinomial theorem to answer this question. Suppose we have a trinomial (a + b + c)ⁿ, we can determine any term to be:
[n!/(n-m)!(m-k)!k!] a^(n-m) b^(m-k) c^k
In this problem, the variables are: x=a, y=b and z=c. We already know the exponents of the variables. So, we equate this with the form of the trinomial theorem.
n - m = 2
m - k = 5
k = 10
Since we know k, we can determine m. Once we know m, we can determine n. Then, we can finally solve for the coefficient.
m - 10 = 5
m = 15
n - 15 = 2
n = 17
Therefore, the coefficient is equal to:
Coefficient = n!/(n-m)!(m-k)!k! = 17!/(17-5)!(15-10)!10! = 408,408