A college student borrows $3,000 for 3 months to pay for a semester of school. If the interest is $61.88 find the monthly paym
2 answers:
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Answer:
So then the best option is:
a. 7.32
Step-by-step explanation:
Previous concepts
Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".
The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"
If we assume that we have groups and on each group from
we have
individuals on each group we can define the following formulas of variation:
And we have this property
Where SST represent the total sum of squares.
The degrees of freedom for the numerator on this case is given by where k =3 represent the number of groups.
The degrees of freedom for the denominator on this case is given by .
And the total degrees of freedom would be
From the info given we know that
And
From definition the F statisitc is defined as:
So then the best option is:
a. 7.32
Assuming we're looking for a power series solution centered around 
, take
Substituting into the ODE yields
The first series starts with a constant term; the second series starts at
; the last starts at 
. So, extract the first two terms from the first series, and the first term from the last series so that each new series starts with a term. We have
Re-index the first sum to have it start at
(to match the the other two sums):
So now the ODE is
Consolidate into one series starting:
![\displaystyle2a_2+(6a_3-a_0)x+\sum_{n\ge1}\bigg[(n+3)(n+2)a_{n+3}+(n-1)a_n\bigg]x^{n+1}=0](https://tex.z-dn.net/?f=%5Cdisplaystyle2a_2%2B%286a_3-a_0%29x%2B%5Csum_%7Bn%5Cge1%7D%5Cbigg%5B%28n%2B3%29%28n%2B2%29a_%7Bn%2B3%7D%2B%28n-1%29a_n%5Cbigg%5Dx%5E%7Bn%2B1%7D%3D0)
Suppose we're given initial conditions
and 
(which follow from setting in the power series representations for 
and 
, respectively). From the above equation it follows that
Let's first consider what happens when
, i.e. 
. The recurrence relation tells us that
and so on, so that
except for when 
.
Now let's consider
, or 
. We know that 
, and from the recurrence it follows that 
for all 
.
Finally, take
, or 
. We have a solution for 
in terms of 
, so the next few terms (
) according to the recurrence would be
and so on. The reordering of the product in the denominator is intentionally done to make the pattern clearer. We can surmise the general pattern for as
So the series solution to the ODE is given by
Attached is a plot of a numerical solution (blue) to the ODE with initial conditions sampled at
and 
overlaid with the series solution (orange) with 
and 
. (Note the rapid convergence.)
Substituting into the ODE yields
The first series starts with a constant term; the second series starts at
Re-index the first sum to have it start at
So now the ODE is
Consolidate into one series starting
Suppose we're given initial conditions
Let's first consider what happens when
and so on, so that
Now let's consider
Finally, take
and so on. The reordering of the product in the denominator is intentionally done to make the pattern clearer. We can surmise the general pattern for
So the series solution to the ODE is given by
Attached is a plot of a numerical solution (blue) to the ODE with initial conditions sampled at