I believe that it is the second and third boxes
More card math! Easy!!
Exactly 1/4 of a 52-card deck of cards is clubs. Another 1/4 spades, another 1/4 is hearts, and the final 1/4 is diamonds. To find what one fourth of 52is, you can divide 52 by 4. That equals 13. So there are 13 clubs in the entire deck of cards. If you are trying <em>not </em>to get a club, you have to find the remainder of the cards without those 13. 52 - 13 = 39. This means that you have 39/52 odds of <em>not </em>getting a club. Simplified you get 3/4.
I did it the long way. A shorter was is to see that, obviously, one whole minus one fourth is three fourths.
Answer:
all work shown and pictured
Answer:
The speed is 16 kilometers per hour.
The cyclist will travel 80 kilometers in 5 hours.
Let
![I = \displaystyle \int e^{-2x} \cos(2x) \, dx[/]texIntegrate by parts:[tex]\displaystyle \int u \, dv = uv - \int v \, du](https://tex.z-dn.net/?f=I%20%3D%20%5Cdisplaystyle%20%5Cint%20e%5E%7B-2x%7D%20%5Ccos%282x%29%20%5C%2C%20dx%5B%2F%5Dtex%3C%2Fp%3E%3Cp%3EIntegrate%20by%20parts%3A%3C%2Fp%3E%3Cp%3E%5Btex%5D%5Cdisplaystyle%20%5Cint%20u%20%5C%2C%20dv%20%3D%20uv%20-%20%5Cint%20v%20%5C%2C%20du)
with

Then

Integrate by parts again, this time with

so that
