
has critical points wherever the partial derivatives vanish:


Then

- If
, then
; critical point at (0, 0) - If
, then
; critical point at (1, 1) - If
, then
; critical point at (-1, -1)
has Hessian matrix

with determinant

- At (0, 0), the Hessian determinant is -16, which indicates a saddle point.
- At (1, 1), the determinant is 128, and
, which indicates a local minimum. - At (-1, -1), the determinant is again 128, and
, which indicates another local minimum.
Try Photo math i guess answer
So, what we need to do is find out the chance you will pull an M or an L out of the scrabble bag on your next turn.
Our first step is finding out how many pieces total are in the bag. This will become the denominator in our answer. To do this, we just need to add up all the pieces we know are in the bag! From the question, we know there are <span>5 As, 3 Es, 1 Z, 2 Ms, 3 Ls, and 1 <span>Y left in the bag. So 5 + 3 + 1 + 2 + 3 +1 should give us 15 total pieces to pick from.
Next, we need to know the total of Ls and Ms left in the bag that we want to pick. This number will be the numerator in our answer. From the question, we know there are 2 Ms and 3 Ls in the bag. Because 2+3 = 5, that means out final fraction for this problem should be 5/15!
Unfortunately, that is not an actual answer for the question, so that means we have to simplify by finding the biggest number that goes into both the top and bottom of our fraction. To get 5, we can only use the numbers 1 and 5. To get 15, we can use 1, 3, 5, and 15. From this, it looks like both the top and the bottom are divisible by 5. When we divide the top by 5 we end up with a 1, and when we divide the bottom by 5 we end up with a 3, meaning our final fraction is D) 1/3!</span></span>
Answer:
If the null hypothesis is true in a chi-square test, discrepancies between observed and expected frequencies will tend to be small enough to qualify as a common outcome.
Step-by-step explanation:
Here in this question, we want to state what will happen if the null hypothesis is true in a chi-square test.
If the null hypothesis is true in a chi-square test, discrepancies between observed and expected frequencies will tend to be small enough to qualify as a common outcome.
This is because at a higher level of discrepancies, there will be a strong evidence against the null. This means that it will be rare to find discrepancies if null was true.
In the question however, since the null is true, the discrepancies we will be expecting will thus be small and common.
Answer:
d = 90°
e = 147°
f = 33°
Step-by-step explanation: