All Answers 358501 Answers

What do you think? 

Is there a huge gap among the data points? 

IF you were given the first 3 points, do you think you could predict the 4th one? What about the 5th one?

How did you get that?

Also, that wasn't correct :(

Wrong answer.

That answer is incorrect.

The average velocity is 1.43 m/s

r^2 = 40/3.

The area of the circle is 24*pi.

Angle ACB = 88 degrees.

We are not here to do your homework for you.

xy = 7/3

There are \(6^4 = 1296\) ways to roll the dice. 

Use stars and bars to count the number of ways to roll a 12. Each roll must be at least 1, so we subtract \(4 \times 1 = 4\) to get 8 stars and \(4 - 1 = 3\) bars.

This makes for \({8 + 3 \choose 3} = {11 \choose 3} = 165\) ways to roll a 12. But, we can't have cases like 6, 1, 1, 0; 6, 2, 0, 0; 7, 1, 0, 0; and 8, 0, 0, 0. 

There are \({4! \over 2!} = 12 \) ways for each of the first 3, and there are \({4! \over 3!} = 4 \) ways for the last one, which sums to 40. 

So, there are 165 - 40 = 125 ways to roll a 12, which makes for a probability of  \(\color{brown}\boxed{125 \over 1296}\) 

There are \({10 \choose 5} = 252\) ways to choose the groups. 

If Jenny and Kenny are in the same group, there are 3 spots left, which makes for \({8 \choose 3} = 56\) ways to be in the same group. 

So, there are 252 - 56 = 196 ways for Jenny and Kenny to be in different groups. 

The probability is then \({196 \over 252} = \color{brown}\boxed{7 \over 9}\)

You drove 90 miles.

The answer is 16 minutes

S = sum_(n=1)^∞ (-1)^(1 + n)/(8 n^3) = (3 ζ(3))/32≈0.112692834671212

T = sum_(n=1)^∞ (-1)^(1 + n)/(-1 + 2 n)^3 = π^3/32≈0.968946146259369

S / T = 0.112692834671212 / 0.968946146259369

==0.11630453880875

Let us call the circle's center \(O\). We first note that if \(A\) and \(B\) are points on the circle, then triangle \(AOB\) is isosceles with \(AO=BO\). Therefore, if \(AOB\) is an obtuse triangle, then the obtuse angle must be at \(O\). So \(AOB\) is an obtuse triangle if and only if minor arc \(AB\) has measure of more than  \(\dfrac{\pi}{2}\)\((90^{\circ})\).

Now, let the three randomly chosen points be \(A_0\),\(A_1\), and \(A_2\). Let \(\theta\) be the measure of minor arc \(A_0A_1\). Since \(\theta\) is equally likely to be any value from \(0\) to \(\pi\), the probability that it is less than \(\dfrac{\pi}{2}\) is \(\dfrac{1}{2}\).

Now suppose that \(\theta<\dfrac{\pi}{2}\). For the problem's condition to hold, it is necessary and sufficient for point \(A_2\) to lie within \(\dfrac{\pi}{2}\) of both \(A_0\) and \(A_0\) along the circumference. As the diagram below shows, this is the same as saying that \(A_2\) must lie along a particular arc of measure \(\pi-\theta\).



The probability of this occurrence is \(\dfrac{\pi-\theta}{2\pi}=\dfrac{1}{2}-\dfrac{\theta}{2\pi}\), since \(A_2\) is equally likely to go anywhere on the circle. Since the average value of \(\theta\) between \(0\) and \(\dfrac{\pi}{2}\) is \(\dfrac{\pi}{4}\), it follows that the overall probability for \(\theta<\dfrac{\pi}{2}\) is \(\dfrac{1}{2}-\dfrac{\pi/4}{2\pi}=\dfrac{3}{8}\).

Since the probability that \(\theta<\dfrac{\pi}{2}\) is 1/2, our final probability is \(\dfrac{1}{2}\cdot\dfrac{3}{8}=\dfrac{3}{16}\).

The value of the first fraction is 3/4.

The smallest possible integer is 14.

How many flavours are there?  Try proof reading your questions.

 a + b = 7,   

a^3 + b^3 = 44, solve for a, b

a≈3.5 - 1.41 i  and   b≈3.5 + 1.41 i

a≈3.5 + 1.41 i ∧ b≈3.5 - 1.41 i

1 / [3.5 + 1.41 i]  +  1 / [3.5 - 1.41 i] =70,000 / 142,381 = 0.4916386315589861.......etc.

With empty boxes allowed, you have:
4 0 0 = 3 permutations
3 1 0 = 6 p
2 2 0 = 3 p
2 1 1 = 3 p

Total = 3 + 6 + 3 + 3 = 15 ways

12! + 15! + 18! + 21! + 24! = 12!(1 + 15...13 + 18...13 + 21...13 + 24...13)

Each of the four products within the brackets is a multiple of 3, so their sum will be a multiple of 3, meaning that the sum of the five numbers within the brackets will be one more than a multiple of 3 and so will not contain 3 as a prime factor.

Any 3 has to come from the 12!

So, 3 -> 1, 6 -> 1, 9 -> 2, 12 -> 1, 

five in all.