All Answers 358087 Answers

Yes, because it can be written as 33/10, the ratio of two whole numbers. 

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x² + y² = 16 is the equation of a circle of radius 4 centred on (0, 0)

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You should get 6.5 Newtons.  (2.5kg*2.6m/s² = 6.5 kg.m/s² = 6.5N)

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yeah

x=4

y=4

or

x=2

y=4

Gotta be careful with how the question is worded titanium.

Probability of getting more heads than tails in a total of 8 flips is the probability of getting 5 heads, 6 heads, 7 heads, or 8 heads.

Almost identical problem to this: http://web2.0calc.com/questions/a-fair-coin-is-flipped-7-times-what-is-the-probability-that-at-least-5-of-the-flips-come-up-heads

Your final answer should be about 0.363

1: Multiply both sides by log(x)

--> (1 * log(x)) / log(x) = (5*log(x)) / 3

--> 1 = (5*log(x)) / 3

2: Multiply both sides by 3

--> 3 = (5*log(x)) * 3 / 3

--> 3 = 5*log(x)

3: Divide both sides by 5

--> 3/5 = (5*log(x)) /5

--> 3/5 = log(x)

4: Assuming your log is to the base e, raise both sides by e

--> e^(3/5) = e^(log(x))

-->e^(3/5) = x .  Something like 1.822.

Use the binomial distribution where N=7 , P = 0.5, 1 - P = 0.5

Calculate the probability of getting 5 heads, 6 heads, 7 heads and then add them all up.

eg. probability of getting 5 heads is (7C5) x (0.5^5) x (0.5 ^2) = 0.164

how come im still alive then? huh?

It's already in its simplest form. If you want to write it in another way though you could express it as sqrt(9^5)

Rome is dead

mmmmm, probably 1/2 cuz its only heads and tails, so 8 coin flips divided by the 2 sides (in this case, heads and tails)=4 and half of 8 as you know is 4, so the probability overall is 1/2

-Titanium Rome

15 miles:6 hours.........................................simple. divide 6 from both sides. 6 divided by 6=1 hour. 15divided by 6=2.5.

Overall, Peter walks 2.5 miles per hour.

-Titanium Rome

Also, zegroes, only ladies can beat men. these women has a secret that we men don't know, and for some reason this makes them stronger than men. ladies beat men, just admit it dude. however, there is one thing: LADIES AND MEN CANNOT DEFEAT BULLETPROOF ROMANS!!!

-Titanium Rome

well guess what, ladies, and gentlemen, step aside. We all know that only Leonardo da Vinci can solve this problem, but I have similar skills in macro photography, so I can probably do this.

Son (anonymous), cubed is cubed. 3 is to the third power for there are cubes. this is used mainly in volume, for its length times width times height. and probably as well has some other benefits. as usual, only people with leonardo da vinci or people with the secret formula can do a problem like this.

-Titanium Rome

P.S. I for now on always make red and black text cuz it represents the best movie ever-Battle Rpyale!!!

oh! I see. Thank you so so much!!

The expected value is the average. So if we have 6 slips of paper with 1 and 2 slips with 3, to get the average we add up all the numbers and divide by the total number of slips, which would give us:

average = (1 + 1 + 1 + 1 + 1 + 1 + 3 + 3)/8 = 12/8 = 3/2

But notice that there are 6 lots of 1 and 2 lots of 3, so we could write this as (6*1 + 2*3)/8 

The 6 and the 2 are known as "weights".  

Notice also that the total number of slips is the same as the sum of the "weights": 6 + 2 = 8.

Whenever you have grouped data (that is, a number of slips with the same value) it tends to be easier to calculate the average by multiplying the "weight" by the number in each group, adding up the results of all the "weights"*numbers and then dividing the whole lot by the sum of all the weights.

Does this make it any clearer?

x is probably 4 cm as well

ps. these pictures creep me out lol XD

You are right Mellie.  I did this one too quickly (as I often do!).  I've now corrected it.

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I think I'm coming round to agreeing with Mellie!  I interpret the question differently each time I read it!

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hi, I was the one that posted this. May you answer the other questions that came along with it please?

$$The probability that the MegaBall matches is $\dfrac{1}{27}$ . The probability that the 5 WinnerBalls match is $\dfrac{1}{\binom{44}{5}}$. So my chances of winning are $\left(\dfrac{1}{27}\right)\times\left(\dfrac{1}{\binom{44}{5}}\right) = \boxed{\dfrac{1}{29,\!322,\!216}}$.$$

Hi, thanks Alan again!

$$$\binom{6}{3}=20$ ways to choose 3 islands. For each of these choices, there is a probability of $\left( \frac{1}{4} \right)^3 \left( \frac{2}{3} \right)^3$ that the chosen islands have treasure and the remaining ones have neither treasure nor traps. Therefore, the probability that the pirate encounters exactly 3 islands with treasure and none with traps is $20 \left( \frac{1}{4} \right)^3 \left( \frac{2}{3} \right)^3 = \boxed{\frac{5}{54}}$.$$H

just tell me what it is?

Is he floating?

Oh my goodness!

What is this madness?