helperid1839321

Looking at this problem, I see an interesting way to apply an integral (in some way) to this situation.

I started approaching this by thinking about how a triangle works. For its three sides to connect, no side can have a length equal to or greater than the sum of the other two sides.

This is the main rule we will use to determine what values are possible.

Now, how do we apply this? Let's think of the information we have. In this case, it's only one piece of information: one of the sides has length two.

Knowing this, we can start to use this in conjunction with the rule we have for a triangle.

x + y >= 2
|x - y| <= 2

If we add in our other limiting factors, we can make an enclosed area, from which we can calculate the area of!

0 < x < 3
0 < y < 3
With all of this, we end up with a shape whose vertices connect in the following order:
(0, 3), (0, 2), (2, 0), (3, 0), (3, 1), (1, 3), (0, 3)

While it's possible to locate the berices on paper by hand, we can also split this hexagon up into three shapes.

For x = (0, 1): (0, 3), (0, 2), (1, 1), (1, 3), (0, 3) [a right trapezoid, with area = 1.5]
For x = (1, 2): (1, 1), (2, 0), (2, 2), (1, 3), (1, 1) [a parallelogram, with area = 2]
For x = (2, 3): (3, 0), (2, 0), (2, 2), (3, 1), (3, 0) [another right trapezoid, with area = 1.5]

Now, we only have to sum up the area this shape takes up, and divide by the area enclosed by x = (0, 3), y = (0, 3) (a 3x3 square, with area = 9)

1.5 + 2 + 1.5 = 5
5/9 = 55.5555...%

I'm actually not to familiar with statistics but would be very happy to learn where you learned this, do you think you'd be able to point me to a resource?

I decided to verify this with programming and found that my program gave the answer as 974. I didn't include details about distribution with that, but a little more poking around with it revealed that the program only found 188 instances of "threes" to your 189. I don't blame you for getting that wrong, counting is hard

Kind of want to do this calculation myself so I'll hop right on board.

The first card we get can be any card in the deck. Let's assume we get the Ace first, or it won't be a blackjack. (This is for purpose of demonstration, and an optimization will be shown later.)

The odds of getting the card we need are 4 in 52, or 1/13.

Now we have 51 cards that we do not know. As long as we don't know what they are, we can say it does not matter if others have been dealt out.

In this case, the odds of getting the 10 card we need, are 16 in 51. 

So, the odds of this are 1/13 * 16/51 = 16/663.

Now, let's go over to the inverse: getting the 10 card first. In this case the odds are 4/13, then 4/51.

In this case, we still get 16/663.

... wait a second... they're the same probability?
Well of course! This isn't a permutation based chance, it's a combination based one. Combination based chances will generally be commutable, and can be shortcut in a few ways. I don't know how to write one for this, but because we have 2 permutations ("2 pick 2") of this, we could have just mulitplied the first answer by 2.

Thus, the odds are 32/663.

Glad this isn't a question of trapezoid perimeter, those are a complete nightmare.

Anyways, let's start by making systems of equations from what we get in the problem.

\(B_L = 4B_S \\ H = (B_L + B_S) \div 2 \\ A = 100\) 

This is what we can get from the description of the problem. Now let's apply what we know about trapezoids to add the important equation that puts it all together:

\(A = \frac{B_L + B_S}{2} \times H\)

"The area of a trapezoid is equal to the average of its bases times its height."

Looks a bit like the second equation above, right?
Let's use some reverse substitution:

\(A = H \times H \\ 100 = H^2 \\ H = 10\)

Now, time for more substitution to solve for the smaller base, as it will result in less shenanigans.

\(10 = (B_L + B_S) \div 2 \\ 20 = 4B_S + B_S \\ 20 = 5B_S \\ B_S = 4\)

And now we solve for the long base:

B_L = 4*4
B_L = 16

Thus, the length of the long base is 16 yards.

Hello Guest, it appears that the numbers in the question failed to render, would it be possible to repost with updated numbers?

Let's first find the ways that we could get products of 90.

One of the obvious first choices would be 6, as it's a factor of 90 and reduces it the most greatly. Let's start with it.

90/6 = 20

Now we can factor 20 into 5 and 4, so we now know that the only valid combination is 6-5-4.

We need the total number of permutations where we get a 6-5-4. In this case, our set is {6 5 4} and we are picking 3. This could be written as "3 pick 3", which happens to be equal to 3!. (3 factorial, not just an excited 3 :) )

Now let's put this over the total number of perumations of dice:

\(\frac{3!}{6^3} = \frac{6}{216} = \frac{1}{36}\)

Thus, the chances of the dice having a product of 90 is 1/36.

This probability problem is simple enough that it seems directed towards a brute force method, which is what I will use.

Let's first write out all the ways that team A could win the tournament, then sum the probabilities.

By games that would have been won:

A-B-A

B-A-A

A-A

You might be scratching your head at only having 2 games with the A-A permutation, but in this case, the third game doesn't matter, and might not be played in some tournament formats. Thus, we only need to calculate for until team A wins.

A-A = 60% * 60% = 36%

A-B-A = 60% * 60% * 40% = 36% * 40% = 14.4% (substitution from A-A chance calculation)
B-A-A = 14.4% by the commutative property of multiplication, it's the same as A-B-A

36%

+ 14.4%

+ 14.4%

=======

64.8%

Therefore, Team A has a 64.8% chance of winning the tournament.

The primary focus of a word problem is teaching you how to turn words into systems of equations, so I'm only going to do that, and if you need furthrer help with the problem, simply ask for the answer and I will follow up with the steps to it.

The first sentence names the perimiter of the rectangle. Let's make a variable for it, P. We're also provided the value of 96 inches. Let's make add this o the system.

P = 96

The second part of the sentence names a relationship between length and width, so let's add variables for those, and call them l and w. We also get the relationship between these two:

l (the length) = 5w ("five times the width") - 6 ("six less than")

This doesn't help us on its own, but there's a third equation that's implied within the problem. The word that tells us is the word rectangle. It doesn't help on its own, but it helps when it adds the relationship between the constant within the problem, and the other relationship between length and width.

Rectangles have the property of their perimeters being twice the sum of their length and height. In equations, for a rectangle:

P = 2*(l + w)

Think you can take it from here?

I would guess that our senses of humor differ notably then, as I wasn't able to detect much humor or anything to laugh at in the post. In any case, the point of not having the original question in the repost still stands.

Assuming we can't choose the same 2 points:

We have 3 points in the same column, or 4 points in the same row, that are valid points to choose, that would result in the dots making a row or column.

Out of the 19 other points you can choose after the first one, there are 7 that match the initial condition.

Therefore, the probability is 7/19.

Sorry I don't have a better explanation, but I'm not really a fan of using latex to make graphics.
... You know what, I'll do it anyways

Here's an example. X is the point we choose, C are the dots that make a vertical line, and R are the ones that make a horizontal line.

\(\begin{bmatrix}X & R & R & R & R \\ C &&&&\\ C &&&&\\ C &&&&\end{bmatrix}\)

Not sure why you would go through the effort of pressing enter so many times, but I might as well put a full walkthrough here.

The guest is correct, the critical information here is that C + D + E = 180, along with the relations.

So, let's expand it through substitution.

4x - 16 + 6x - 1 + 4x - 13 = 180

We can also group the terms using the commutative property, making this a lot easier to look at: 

14x - 30 = 180

Now, we solve a single-variable equation.

14x = 210

7x = 105

x = 15

.

Now, let's plug this value of x back into the equation for angle C.

C = 4x - 16
C = 4*15 - 16

C = 60 - 16
C = 44

Thus, angle C has a measure of 44 degrees.

And guest, I like the fact that you didn't give the answer right away but instead gave a hint to solve the problem, that's better than a number of other guest posts I've seen on here as answers :D

I've already mentioned this to another user, but simply posting a link without the original question is not generally seen as a productive way to attract an answer. If the question has been buried and needs an answer, general policy is that it's grounds to repost the orginal question, optionally with a link to the original post, as far as I remember.

The title isn't necessarily helpful either, as it doesn't indicate what the problem was. It's quite possible not many people saw it or had someone else to help or something else to do. We don't trawl this forum all day and appreciate having other things we can do with our lives.

Just a kind message asking you to please refrain from doing this in the future, thank you for listening.

This one appears solvable without graphing or using a calculator due to one quirk:
Take another look at those equations for the circles

(x-2)² + (y-3)² = 9

(x-2)² + (y+4)² = 16

The x-term in each equation is actually identical. We can therefore eliminate each from the equation for the purposes of solving for y.
We'll also expand the y-term here for convenience in the next step.

y² - 6y + 9 = 9
y² + 8y + 16 = 16

We can also eliminate the term y² here as it's shared on both sides. We can also take out the constant values on both sides, giving us...

-6y = 0

8y = 0

Through multiplication of each equation we can get y = 0. Now plugging this into the original equations...

(x-2)² + (0-3)² = 9

(x-2)² + (0+4)² = 16
 

(x-2)² + (-3)² = 90

(x-2)² + (4)² = 16

(x-2)² + 9 = 9

(x-2)² + 16 = 16

(x-2)² = 0

(x-2)² = 0

Taking square root of both sides...

x-2 = 0

x-2 = 0

x = 2

x = 2

Going back, we now have 2 solutions for (2, 0). Thus, the question is worded misleadingly, and the only intersection is at (2, 0)

Let's make a system of equations to express the system to make it easier to not make a mistake (as the guest answerer unfortunately did).
 

m = max's baseball cards
p = pauls baseball cards

5p = m
p + 28 = m
 

... using substitution of the term m...
 

5p = p + 28

4p = 28
p = 7

... solve for m...

5p = m
5*7 = m
m = 35

Therefore, Max has 35 baseball cards.

I'm going left to right with the multiplication. Q is across the top, E is down the left.\(\left.\begin{matrix} &11&-5i\\ 3&33&-15i\\ -2i&-22i&-10 \end{matrix}\right] = 23 - 37i\).

Now, we multiply QE by D. \((23-37i) \times 2i = QE \times D = QED = 46i + 74\).
 

Remember, i * i = -1, and -i * i = 1. i is effectively a variable until you are dealing with more than one of them being multiplied.

I think the term for this is "abstract quantity"? It's a value you use like a variable with some exceptions.

This response was not very productive, and does not provide me with any helpful feedback to use to improve. Feedback with the objective of helping people grow would be appreciated more than a response with little thought, depth, or room to grow from. Thank you :)