supermanaccz

198 is your answer

We cannot see the graph

huh?

you lost me

Can someone please help me?

THis is due tmr and I have no clue

I don't understand your question, can you repost in proper form?

Here is your answer Guest- 0.06, 0.603, 0.633, 0.6, 0.63

Have fun with your answer

Here is my solution: 

To draw two cards we can do $\binom{52}{2}$, which is 1326. 

Since there are 4 aces in the deck, the number of ways we can choose 2, is $\frac{\binom{4}{3}}{2}$, which is 6. 

6/1326 = 1/221.

So the answer is D, 1/211; no, they are dependent events. 

This is a system of equations problem.  

To easiest way to solve this is first add both equations. 

12x + 21y + 21x + 12y = 51 + 15

So 33 y + 33x = 66 

x + y = 2 

We can use x = 2 - y and plug this in to equation #1.

12 ( 2 - y ) + 21y = 15 

24 - 12y + 21y = 15

y = -1 

x = 3 

The ordered pair you are looking for is (3, -1)

We can already plug in 78 for d in this equation. 

78 = 0.05s^2 + 1.1s 

After subtracting 78 from both sides, we have:

0.05s^2 + 1.1s - 78 = 0.

Solving this quadratic, we have: s = 30.0000000000000004, -52.00000000000001

The last time I checked, we can’t go negative speed, so s = 30

It is actually 30.000000000000004, but just round it down. 

Awww...

My latex is not showing up:(