Homework Statement
The amounts of time that a cashier spends processing individual customers' orders are independent random variables with mean 2.5 minutes and standard deviation 2 minutes.
a) What is the approximate probability that it will take more than 4 hours to process orders of 100...
Homework Statement
The number of claims that an insurance company receives per week is a random variable with the Poisson distribution with parameter λ. The probability that a claim will be accepted as genuine is p, and is independent of other claims.
a) What is the probability that no claim...
The gravitational force and normal force from the earth do not cancel out though. There is a reaction force from the Earth equal in magnitude to the force applied to the ball by the man. This force is not cancelled out by the force of the cart pushing on the Earth because the Earth is not in...
Say that you take the Ball, Man, Cart system. The forces between the ball and the man, and the man and the cart are all action-reaction pairs and thus cancel out. But the cart also pushes on something outside of the system: the Earth. The Earth pushes back with a reaction force. Is this force...
Perhaps I'm misunderstanding the question, but if a skier goes off a jump with a known velocity, then their path through space is determined. Isn't it clear that he can only land in one particular spot on the ramp?
I think it's safe to assume that the useful energy in this case is the light being emitted from the bulb. Which would mean that 51 J of light energy is emitted. The other energy doesn't "remain in the circuit" it is lost to heat.
Not quite. The horizontal part of the triangle is her resultant velocity,or the vector sum of her speed with respect to the water and the current flow with respect to the shore. Her velocity relative to the water should be inclined at some angle with the horizontal, making it the hypotenuse of...
This is correct.
I think you may be overthinking this. If she ends up 15.36 m downstream, that's how far she will have to walk to reach the market.
Your intuition is right and you are correct that this is basically a trigonometry problem. Knowing that her speed relative to the water is 2.5...
Ok, so assuming we reduce the set in U to a basis: then if v is in U the given generating set for W is linearly dependent. Then v would be removed from this set, making it linearly independent. This implies that U and V share a basis, and thus their dimensions are the same. If v is not in U...
Homework Statement
Given v1, v2 ... vk and v, let U = span {v1, v2 ... vk} and W = span {v1, v2 ... vk, v}. Show that either dim W = dim U or dim W = 1 + dim U.
The Attempt at a Solution
I'm not really sure where to start. If I knew that {v1, v2 ... vk} was linearly independent, then it would...
Thanks for the reply! Today I was able to prove (1) by contraposition but I thought that meant I had to prove (2) by contraposition as well. Can I just prove the positive statement for (2)?
Edit: Wait, taking statement A as {p, q, pq} is linearly independent and statement B as deg p ≥ 1 and deg...
Homework Statement
Let {p, q} be linearly independent polynomials. Show that {p, q, pq} is linearly independent if and only if deg p ≥ 1 and deg q ≥ 1.
Homework Equations
λ1p + λ2q = 0 ⇔ λ1 = λ2 = 0
The Attempt at a Solution
λ1p + λ2q + λ3pq = 0
I know if λ3 = 0, then the coefficients of...
Homework Statement
Is U = {A| A \in nℝn, A is invertible} a subspace of nℝn, the space of all nxn matrices?
The Attempt at a Solution
This is easy to prove if you assume the regular operations of vector addition and scalar multiplication. Then the Identity matrix is in the set but 0*I and...
Do you know what I did? I simplified the original function to get the one I posted here, just checked and they're not the same. That's why the derivative didn't match up. :surprised I think I've about got it sorted out now.
Somebody want to help with this derivative:
y = (7.75x3 + 2250) / 750x
I differentiate it with the quotient rule and get:
dy/dx = 31x / 1500 - 3 / x2
But that's wrong. It's got a zero around 5 - 6 and the original function has its minimum at (4.03, 1.12). Not sure what I did wrong.
No, I multiplied that in at the very beginning. I scaled up the fraction by 1000 times so I had a 4 in the numerator rather than a decimal. After fixing what Reptilian mentioned, my derivative looks correct, and the zero is at the x-value of that local minimum, so I think it's right.
Using the quotient rule on the 4∏r2 / 3000 term. The denominator of the differentiated expression is 16∏2r4. A pi and an r cancel from the numerator which was 8∏r(3000).
Yeah I noticed. No worries. :tongue: Do you know what's wrong with the derivative. I found just with the power rule and the quotient rule on the last term.
After simplifying I get:
(0.003 + 0.004√2) ∏r2 + 3/r - (4∏r2) / 300
Which is equivalent now according to a graphing calculator.
Differentiating, I get:
(0.006 + 0.008√2) ∏r - 3r-2 - 1500/∏r3
But thats not right. i.e. it has positive values when the slope of the function is negative. :/
Does somebody want to help me with differentiating this function:
(0.003 + 0.004\sqrt{2})∏r2 + 0.004∏r [(2250 - ∏r3) / 3∏r2)]
Originally I tried simplifying it by distributing the 0.004∏r into the fraction, but whenever I did that the two expressions were never equivalent when I checked them...
Take the two equations I've quoted, and set them equal to each other. The stones have the same displacement:
1/2at1^2 = 1/2at2^2 + vi * t2
Now sub in a = 9.8 m/s2, t2 = t1 - 1.50 s, vi = 24.0 m/s, and leave t1 as the unknown. The squared terms can be subtracted out and you are left with a...
The equation for stone number one is just
d = 1/2at1^2
Stone 2:
d = 1/2at2^2 + v0 * t2
Remember that t2 is t1 - 1.50 and that the displacements of the two stones are equal.
How do you determine the equation of all possible tangents to a curve (say, a parabola) that pass through a given point that is not on said curve? This is more of a conceptual question, and it's not homework, so I thought it fit in this forum. I think there might be a question like this on the...
Well, when the electron is accelerated through two plates with a potential difference of 26000V, the electron has lost 26000V of electric potential. What does this tell you about the electric potential energy of the electron at the moment it exits the plates? In turn, what does that tell you...