$\int {( \sqrt[4]{ x^{3} } + \sqrt[3]{ x^{2} }) } \, dx = \\ = \int {( x^{3/4}+ x^{2/3} )} \, dx = \\ \frac{ x^{7/4} }{7/4}+ \frac{ x^{5/3} }{5/3}+C = \\ \frac{4 \sqrt[4]{ x^{7} } }{7}+ \frac{3 \sqrt[3]{ x^{5} } }{5}+C$

4 0

2 years ago

Two six-sided fair dice are rolled. The probability that at least one number is odd and the sum of the two numbers is even is .

Oxana [17]

The easiest way to figure out probability problem with small data sets is to write out your entire sample space then divide by the total:

Sample size = 6 * 6 = 36

S = {[1,1],[1,2],[1,3],[1,4],[1,5],[1,6],[2,1],[2,2],[2,3],[2,4],[2,5],[2,6],[3,1],[3,2],[3,3],[3,4],[3,5],[3,6],[4,1],[4,2],[4,3],[4,4],[4,5],[4,6],[5,1],[5,2],[5,3],[5,4],[5,5],[5,6],[6,1],[6,2],[6,3],[6,4],[6,5],[6,6]}

The only way to make a number combination that's even while 1 die is odd is to have 2 odd numbers.

{[1,1],[1,3],[1,5],[3,1],[3,3],[3,5],[5,1],[5,3],[5,5]}

This gives us 9 results.

The probability of this happening is 9/36 = 1/4 = 0.25

Now if we have to get a 6 with the product being at most 15 we know that the biggest number that 6 can be multiplied by is 2 which gives us 12.

We are left with 4 options:

{[1,6],[2,6],[6,1],[6,2]}

The probability of this happening is 4/36 = 1/9 = 0.1111...

5 0

2 years ago