The answer in itself is 1/128 and here is the procedure to prove it:
cos(A)*cos(60+A)*cos(60-A) = cos(A)*(cos²60 - sin²A)
<span>= cos(A)*{(1/4) - 1 + cos²A} = cos(A)*(cos²A - 3/4) </span>
<span>= (1/4){4cos^3(A) - 3cos(A)} = (1/4)*cos(3A) </span>
Now we group applying what we see above
<span>cos(12)*cos(48)*cos(72) = </span>
<span>=cos(12)*cos(60-12)*cos(60+12) = (1/4)cos(36) </span>
<span>Similarly, cos(24)*cos(36)*cos(84) = (1/4)cos(72) </span>
<span>Now the given expression is: </span>
<span>= (1/4)cos(36)*(1/4)*cos(72)*cos(60) = </span>
<span>= (1/16)*(1/2)*{(√5 + 1)/4}*{(√5 - 1)/4} [cos(60) = 1/2; </span>
<span>cos(36) = (√5 + 1)/4 and cos(72) = cos(90-18) = </span>
<span>= sin(18) = (√5 - 1)/4] </span>
<span>And we seimplify it and it goes: (1/512)*(5-1) = 1/128</span>
Answer:
uhm 1/4???
Step-by-step explanation:
When there are 1000 repetitions, the mean will be 5.002.
The charts that you have list the number of total heads when you flip the coin 10 times in each trial.
Multiply the total of each by the number number of heads for that category and divide by 1000.
1 x 0
8 x 1
43 x 2
117 x 3
2017 x 4
248 x 5
203 x 6
121 x 7
45 x 8
6 x 9
1 x 10
If you add up those products and divide by 1000, you have 5.002.
Using the law of large numbers, the experiment with 1000 rolls will be the closest to the theoretical amount.
Answer:
Substitute the given n-values into each function, f(n), to determine which generates the same values as the given a(n)'s.
Step-by-step explanation:
