Good luck on your journey. I think you're making a very positive choice. I think a lot of people in your position would be too scared to change their life path because it's kind of drilled into our heads that we should finish school and get settled down asap. I found out myself that life isn't...
I think in terms of increasing retention and also preventing burnout, it makes the most sense to take just one of the mathematics courses.
The material will be condensed but if that's all you're focusing on then you'll have plenty of time to keep up. Also, since you'll be spending a significant...
It sounds like you're not completely sure what is meant by the median or how to find the distribution of it. I want to explain those concepts to you so you can try to tackle the problem.
You have a sample of 5 random variables all distributed uniformly over (0,1). So we have X_1, X_2, X_3...
Homework Statement
Prove or give a counter example of the following statement:
If f: [a,b] \to [c,d] is linear and g:[c,d] \to \mathbb{R} is Riemann integrable then g \circ f is Riemann integrable
Homework EquationsThe Attempt at a Solution
I'm going to attempt to prove the statement is...
Hey there's actually a way you can do it without too much thinking:
y = x^r \\
y' = rx^{r-1} \\
y'' = r(r-1)x^{r-2} \\
\implies xy'' - y' = x[r(r-1)x^{r-2}] - rx^{r-1} = x^{r-1}r(r-1 -1) = x^{r-1}r(r-2)
Which equals zero for all x when r = 0 or r = 2. Sorry I didn't notice this before. A bit...
You kind of have to think a bit outside the box to figure out which values of r are "different".
If r = 0 then y is a constant so all if its derivatives are 0
If r = 2 then its first derivative is a function of x but its second derivative is a constant
if r < 0 or r >2 then all the derivatives...
What you should do is plug in some concrete examples. I'll give you two and see if you can work out the rest:
y = x^1 \\
y' = 1 \\
y'' = 0 \\
\implies xy'' - y' = x(0) - 1 \neq 0
So you can see it doesn't work for r = 1. Now let's see why it does work for r = 2
y = x^2 \\
y' = 2x \\
y'' = 2...
Yep you've got it. Think of the two critical regions as events. If the test statistic falls in the first region OR the second region we reject the null hypothesis. Since The two regions are disjoint, we add the probabilities using the usual probability rules (P(A or B) = P(A) + P(B) if A and B...
Imagine they didn't say anything about vectors and just gave you the equation "2*a + 1.5*b + 1*c", would you know what it describes? You're multiplying the price of something by the number of those things you sold. What does that mean in real life?As for the vectors:
Without all this business...
One thing I noticed right away is that you're confusing the unknown population means with the sample means that can be calculated by doing the survey. The symbol "mu" always refers to the unknown population means while the symbol x with a bar over it represents the mean you get from your sample...
Actually a relatively simple function works lol. Sorry for all this confusion:
h(x) = xf(x)
There exists a c in (0, 1) such that
0 = cf'(c) + f(c)
by Rolle's Theorem.
Hey sorry I edited my previous post. What I wrote doesn't work because the two c's might be different. The c that works for the equation you posted might be a different c from the one I used for h(x).
When trying to define an h(x) we want it to be 0 at the end points so we can apply Rolle's...
edit
Sorry the post I made contained an error. I'm not quite sure how to fix it yet but basically what I think you need to do is find a function h(x) and define it in terms of some linear combination of f(x), f(0), f(1) so that it equals zero at the end points and then apply Rolle's Theorem.
When x is between -1 and 2 one of the functions is lower than the other. When you plot them can you see which it is? That is the lower bound. And the upper bound is the function that is higher.
The upper and lower bound varies with x. That's why you can't just use constants...you need to write...