Alan

Here's a numerical approach:

.So x ≈ 2.162939

.

Here are the steps:

.

As follows:

.

Multiply by √5 divided by √5:

\(\frac{2\sqrt3}{\sqrt5}\rightarrow \frac{2\sqrt3\sqrt5}{\sqrt5 \sqrt5}\rightarrow \frac{2\sqrt{15}}{5}\)

.

Given the way the problem is written Guest#1 is correct!

However, if the problem should have said "Max gave 2/5 of his beans away to his brother, and consumed 1/3 of his remaining beans. He takes the remaining amount of beans and places it evenly into 3 containers. If each container had 2 beans how many beans did he have to start (in the beginning)?", then:

Let B be the original number:

After giving away 2/5B max is left with 3/5B.

After eating 1/3 of this Max is left with (2/3)*3/5B → 2/5B

This equals 2*3 → 6

so 2/5B = 6

B = 6*5/2  → 15

So Max originally had 15 beans.

2 out of 10 digits are greater than 7, so the probability of choosing these is 2/10.  Since there are 25 students the number likely to choose a number greater than 7 is (2/10)*25 or 5.

.

Quote from Wikipedia:

"A t-test is any statistical hypothesis test in which the test statistic follows a Student's t-distribution under the null hypothesis. It can be used to determine if two sets of data are significantly different from each other."

.

The probability of rolling an even number is 1/2

The probability of rolling a 3 is 1/6

Since the probabilities are independent the probability of rolling an even followed by a 3 is just the product of the two, namely (1/2)*1/6) → 1/12

. 

We have:

 \(x = r\times \cos \theta \text{ and }y = r\times \sin \theta \text{ so:}\\ r^2(\cos^2\theta-\sin^2 \theta)=1 \text{ or:}\\r^2\times \cos2\theta=1\)

.

Here's another graph to illustrate the point:

.Seems odd that the "maximum" has a lower value than the "minimum"!!

.