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Hello Guest. This is my best attempt at the problem above. (I'm not very good at probability problems, but I might as well try :D)

"James and Jeff are planning to meet at Andrew’s house at a random time from 6:00 to 9:00. If James arrives before Jeff, then what is the probability James arrives before 7:00?"

SOLUTION:

I'm not so sure why the problem included Jeff. He can just be left out, I believe, since the probability of Jame's arrival time has nothing to do with Jeff's.

With that said, we can calculate the probability of just James arriving before 7:00.

From 6:00 to 7:00, there's a time period of 1 hour, since exactly 1 hour passes. From 6:00 to 9:00, there's a time period of three hours, since 3 hours pass. Therefore, our probability is $\frac{1}{3}$.

Hope this helped! 

Hello Guest.

A factorial, denoted as $n!$, is the same thing as $n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1$. In other words, it is the product of all positive integers between 1 and $n$, inclusive.

There are other variations of this, such as the double factorial ($n!!$), which basically means the product of all even integers between $n$ and 1, inclusive.

I hope this helped! 

area of triangle OBC = 0.245049014 utility squared

The diagonals of a rhombus measure 18 feet and 16 feet. What is the perimeter of the rhombus? Express your answer in the simplest radical form.

Perimeter   P = 4[sqrt(8² + 9²)]

In trapezoid ABCD, sides overline AB and overline CD are parallel, angle A = 2 * angle D, and angle C = 8 * angle B. Find angle A.

∠A = 120º      ∠B = 160º         ∠C = 20º        ∠D = 60º

I believe the answer is $81$.

Awesome thank you so much. This information is hard to find. I guess not too many people are interested in it, because it's a fact that light is always moving and hence doesn't have a rest mass. 

1.  Subtract (p.c)²  from both sides. to get  E² - (p.c)² = (m.c²)² 

2.  Take the square root of both sides to get \(\sqrt{E^2-(p.c)^2}=m.c^2\)

3. Divide both sides by c² to get \(\frac{\sqrt{E^2-(p.c)^2)}}{c^2}=m\)

4. Substitute for E and p using the expressions given.

Honestly, could you just show the entire process, I'm still confused as to how you divided by c^2 without dividing the other side as well. Plus I just don't get where the minus comes from.

Can you explain how you switched the sides of the m, E, and most importantly how you came upon a minus sign for that nifty trick?

Like so

Divide the entire equation by 2    to get  y= mx + b   form of the line.....slope is m

Then perhaps you can demonstrate the multiplication of zero or the deduction of the total sum of it's parts using the formula provided and back up this claim.

But a "photon of light" DOES NOT have any mass!! That is the reason light travel at its well-known speed, which is the "maximum" speed possible. A photon may have "wavelength" or "frequency" and , of course, "energy", but no "mass".

Well as a general rule of thumb i've come to recognize is that to get zero you either deduct the exact sum of the parts or multiply by zero. By my calculations there is no deduction that is happening equal to the sum of the parts and you are certainly not multiplying by zero, so light must have mass. Short answer is i'm trying to figure out the mass of light with a wavelength of 685

It would have been helpful had you proofread your Q and then included the equation of the line ! D'Oh!

This is a sideways parabola....for a vertical line to intersect at only one point , it must just touch the vertex (tangent at the vertex)

    the  y coordinate of the vertex is   -b/2a       - -3/(2*-2) = - 3/4

    now sub this back into the equation to calculate the x coordinate of the vertex....this will be the 'k' value you are seeking....

What "mass" are you trying to calculate? Don't confuse the two "E's", energy equations:

1 - E =hf, where h=Planck's Constant=6.626 x 10^-34 Js. f=Frequency.
f =c/λ, where c=speed of light, λ =Wavelength of the photon.

E =hc/λ  [This is the equation that neasures the energy of a photon].

2 - E =MC^2, where M =Mass, C=speed of light. [This is the famous Einstein's equation, equating the mass(of an object) to its energy].

Thank you for your time.  
I managed to figure this one out...

What is the largest integer n such that \(7^n\) divides 1000!

Alternate form
One may also reformulate Legendre's formula in terms of the base-p expansion of m.
Let s_{p}(m) denote the sum of the digits in the base-p expansion of m.

Given
\(\begin{array}{|rcll|} \hline px+qy+rz&=&1 \\ p+qx+ry&=&z \\ pz+q+rx&=&y \\ py+qz+r&=&x \\ p+q+r &=&-3 \\ \hline \end{array}\)

Find x+y+z.

\(\small{ \begin{array}{|lrcll|} \hline &px+qy+rz&=&1 \\ &p+qx+ry&=&z \\ &pz+q+rx&=&y \\ &py+qz+r&=&x \\ \hline \text{sum} & (px+py+pz) +(qx+qy+qz) \\ & +(rx+ry+rz) + (p+q+r) \\ & &=& 1+x+y+z \\\ & p(x+y+z) +q(x+y+z) \\ & +r(x+y+z) + (p+q+r) \\ & &=& 1+x+y+z \\ & (x+y+z)(p+q+r)+ (p+q+r) &=& 1+x+y+z \\ & (p+q+r)+(x+y+z)(p+q+r) &=& 1+x+y+z \\ & (p+q+r)(1+x+y+z) &=& 1+x+y+z \quad | \quad p+q+r=-3 \\ & -3(1+x+y+z) &=& (1+x+y+z) \quad | \quad -(1+x+y+z) \\ & -4(1+x+y+z) &=& 0 \quad | \quad : (-4) \\ & 1+x+y+z &=& 0 \\ & \mathbf{x+y+z} &=& \mathbf{-1} \\ \hline \end{array} }\)

      θ  = 16π + π/4    as sin(16π) is zero and √2/2 is the opposite/hypotenuse of a 45-45-90 triangle.

As written the only way the question would make sense is if a = 0 and b = 17.  However, like Melody, I suspect there is something missing.

Unfortunately, I really do not understand the logic behind what you have done.

That’s interesting ...if not off-putting. 

What part of my logic do you not understand? You’ve demonstrated logic and mathematics for probability questions far more advanced and complex than this one.  Investing a few minutes of time in analyzing the steps should bring brilliant illumination in understanding the logic.  If there is a flaw in my logic, it should be discernable in one of the steps outlined above. 

Here’s a list of all possible die rolls with all possible combinations of H/T for the three coins. 

\(\begin{array}{|c|c|c|} \hline 1& 6 & HHH\\ 2& 6 & HHT\\ 3& 6 & HTH\\ 4& 6 & HTT\\ 5& 6 &THH\\ 6& 6 &THT\\ 7& 6 &TTH\\ 8& 6 &TTT\\ \hline \end{array} \)  \(\begin{array}{|c|c|c|} \hline 9& 5 & HHH\\ 10& 5 & HHT\\ 11& 5 & HTH\\ 12& 5 & HTT\\ 13& 5 &THH\\ 14& 5 &THT\\ 15& 5 &TTH\\ 16& 5 &TTT\\ \hline \end{array} \)   \(\begin{array}{|c|c|c|} \hline 17& 4 & HHH\\ 18& 4 & HHT\\ 19& 4 & HTH\\ 20& 4 & HTT\\ 21& 4 &THH\\ 22& 4 &THT\\ 23& 4 &TTH\\ 24& 4 &TTT\\ \hline \end{array} \)   \(\begin{array}{|c|c|c|} \hline 25& 3 & HHH\\ 26& 3 & HHT\\ 27& 3 & HTH\\ 28& 3 & HTT\\ 29& 3 &THH\\ 30& 3 &THT\\ 31& 3 &TTH\\ 32& 3 &TTT\\ \hline \end{array} \)   \(\begin{array}{|c|c|c|} \hline 33& 2 & HHH\\ 34& 2 & HHT\\ 35& 2 & HTH\\ 36& 2 & HTT\\ 37& 2 &THH\\ 38& 2 &THT\\ 39& 2 &TTH\\ 40& 2 &TTT\\ \hline \end{array} \)   \(\begin{array}{|c|c|c|} \hline 41& 1 & HHH\\ 42& 1 & HHT\\ 43& 1 & HTH\\ 44& 1 & HTT\\ 45& 1 &THH\\ 46& 1 &THT\\ 47& 1 &TTH\\ 48& 1 &TTT\\ \hline \end{array} \)

Randomly pick a number from 1 to 48, and then do it again. 

To me, the logic and math used for the tabulation of these simultaneous events seems simple and straightforward.  The only complexity, which seems trivial after a full view, comes from the relative time (dimension) introduced in the question by use of the word “before,” What is the probability that I flip three heads twice in a row before I roll two 6's twice in a row?

This necessitates setting up an equation to calculate a probability that one event (two sequential sixes) occurs BEFORE another event (two sequential sets of three-heads) occurs.  This requires factoring out non successes –where neither a six nor a set of three heads occurs, leaving only the successes of the die or the coins. 

After factoring, (28.125%) of the twin events in the sample set will have a success for either the die or the set of coins. There are no ties in this sample set: if the three heads appear simultaneously with a six on the die, twice in two events then the coins have the success because the six has to appear twice BEFORE the set of three-heads occurs. In this space, the coins are successful (44.44%) of the time and the die is successful (66.66%) time. This means randomly sampling from this set will produce a selection where the coins are successful (44.44%) of the time time when compared to the die.

So the probability that a set of 3 coins has three heads twice in sequence before a die has two sixes in sequence is (44.44%).

GA