Answer:
I would go with the second one.
Step-by-step explanation:
Answer:
43.531
Step-by-step explanation:
This problem needs the law of cosines since the three sides of the triangle are involved as well as one angle. The law of cosines is:
c^2 = a^2 + b^2 - 2ab*cos(C)
Where the lowercase letters are side values and capital letters are angle values. Just in case I will mention side a is the one with a length of 14, side b is 20 and c = 12.
Since it is asking for angle A instead of angle C we can rewrite the law of cosines to fit that, basically just rearranging the letters.
a^2 = c^2 + b^2 - 2cb*cos(A)
Now we just plug in and solve.
14^2 = 12^2 + 20^2 - 2*12*20*cos(A)
Rearrange to get A by itself
Now we take the inverse cosine, or arccos of both sides to get our answer.
Let me know if there is something in my explanation you don't understand.
In probability and statistics, there is an equation used for repeated trials. In this equation, you find the probability of getting 'r' successful events out of 'n' trials. Moreover, you should incorporate the probability of the success per trial. For a 6-sided face, each side has a probability of success of 1/6. On the other hand, the probability of each side not appearing is 5/6, because when you add these two, it would sum up to 1 as always.
Now, the equation for repeated trials is:
Total Probability = n!/r!(n-r)! * p^(n-r) * q^r
where
n = 5 tosses
r = 1 and 2 (since you want to see the probability of a side less than 3)
p = 1/6
q = 5/6
So, you add the individual probability when r=1, and when r=2.
Total Probability = 5!/1!(5-1)! * (1/6)^(5-1) * (5/6)^1 + 5!/2!(5-2)! * (1/6)^(5-2) * (5/6)^2
Total Probability = 50/243 or 20.58%
Answer:
Step-by-step explanation:Very interesting problem. Be thankful it is not a six sided dice.Part A1 11 21 31 42 12 22 32 43 13 23 33 44 14 24 34 4Part BThere are 16 possible out comes1 44 12 33 24 out of the 16 outcomes are possibleP(5) = 4/16 = 0.25Part C The theoretical out come would be 4 times4/16 = 0.25On a hundred rolls, you would expect to get 25 sets of doubles.Step-by-step explanation:
Answer:
The first question's answer is c 1.0, 2.0, 3.0
Step-by-step explanation: