Find the slope of line containing the given points. (3,-8),(1,4)
2 answers:
You might be interested in
Answer:
Step-by-step explanation:
Hello!
You need to construct a 95% CI for the population mean of the length of engineering conferences.
The variable has a normal distribution.
The information given is:
n= 84
x[bar]= 3.94
δ= 1.28
The formula for the Confidence interval is:
x[bar]±*(δ/n)
Lower bound(Lb): 3.698
Upper bound(Ub): 4.182
Error bound: (Ub - Lb)/2 = (4.182-3.698)/2 = 0.242
I hope it helps!
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
Answer:
and
Step-by-step explanation:
We are given the term .
<em>It is known that, 'A monomial is an algebraic expression consisting of one term'.</em>
So, we add the given options to .
1.
2.
3.
4.
5.
6.
7.
As, we can see that only option 2 and 6 results in an expression having one term.
Thus, adding and
to
results in a monomial.