Answer:
0.45% probability that they are both queens.
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes
The combinations formula is important in this problem:
is the number of different combinations of x objects from a set of n elements, given by the following formula.
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
Desired outcomes
You want 2 queens. Four cards are queens. I am going to call then A,B,C,D. A and B is the same outcome as B and A. That is, the order is not important, so this is why we use the combinations formula.
The number of desired outcomes is a combinations of 2 cards from a set of 4(queens). So
![D = C_{4,2} = \frac{4!}{2!(4-2)!} = 6](https://tex.z-dn.net/?f=D%20%3D%20C_%7B4%2C2%7D%20%3D%20%5Cfrac%7B4%21%7D%7B2%21%284-2%29%21%7D%20%3D%206)
Total outcomes
Combinations of 2 from a set of 52(number of playing cards). So
![T = C_{52,2} = \frac{52!}{2!(52-2)!} = 1326](https://tex.z-dn.net/?f=T%20%3D%20C_%7B52%2C2%7D%20%3D%20%5Cfrac%7B52%21%7D%7B2%21%2852-2%29%21%7D%20%3D%201326)
What is the probability that they are both queens?
![P = \frac{D}{T} = \frac{6}{1326} = 0.0045](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7BD%7D%7BT%7D%20%3D%20%5Cfrac%7B6%7D%7B1326%7D%20%3D%200.0045)
0.45% probability that they are both queens.
Answer:
-80
Step-by-step explanation:
We have to do the parentheses first because it's part of PEDMAS.
-2x-5= 10
Next, we take 10 times -8
-8 x 10= -80
There you go! Hope this helps!
D 32 because 160 divided by 5 is 32. hope this helped
Answer:
concentration
Step-by-step explanation:
Because concentration is a factor to how much of a volume to give
Answer:
she messed up on step two because she has to subtract 10 from both sides
Step-by-step explanation:
step 1: -6(x+3)+10<-2
step2:-6(x+3)+10-10<-2-10
step3: -6(x+3)<-12
step4: (-6)(x+3)(-1)≥(-12)(-1)
step5:6(x+3)>12
step6:divide both sides by 6
step7:simplify and subtract 3 from both sides and then simplify again