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The mean of 10 observations is 21 If each score is divided by 3 find the new mean?
The new mean would be 7. The mean is the average of the data. (x1+x2+x3+x4+x5+x6+x7+x8+x9+x10)/10=21 [(x1/3)+(x2/3)+(x3/3)+(x4/3)+(x5/3)+(x6/3)+(x7/3)+(x8/3)+(x9/3)+(x10/3)]/10=? [(1/3)(x1+x2+x3+x4+x5+x6+x7+x8+x9+x10)]/10=? [(x1+x2+x3+x4+x5+x6+x7+x8+x9+x10)/10]/3= 21/3=7
What is the pattern for 1 1 2 6 24?
The pattern is x1, x2, x3, x4, x5, x6.... The next two numbers are 120, 720.
How do you solve x2 equals cos x?
You can solve this to the accuracy of your liking by using Newton's method: xn+1 = xn - f(xn) / f'(xn) In this case, we'll say f(x) = x2 - cos(x) f'(x) would then be 2x + sin(x) Let's take a rough guess, and start with x0 = 0.5 x1 = 0.5 - (0.52 - cos(0.5)) / (2(0.5) + sin(0.5)) = 0.92420692729319751536 x2 = x1 - (x12 - cos(x1)) / (2x1 + sin(x1)) = 0.82910575599741780916 x3 = x2 - (x22 - cos(x2)) / (2x2 + sin(x2)) = 0.82414613172819520712 x4 = x3 - (x32 - cos(x3)) / (2x3 + sin(x3)) = 0.8241323124099124229 x5 = x4 - (x42 - cos(x4)) / (2x4 + sin(x4)) = 0.82413231230252242297 x6 = x5 - (x52 - cos(x5)) / (2x5 + sin(x5)) = 0.82413231230252242296 Now we can test our answer: 0.824132312302522422962 = 0.67919406818110235182 cos(0.82413231230252242296) = 0.67919406818110235183 So we're accurate to the nearest ten quintillionth.
Find the number of distinguishable permutations of the letters in the word hippopotamus?
First of all, find the total number of not-necessarily distinguishable permutations. There are 12 letters in hippopotamus, so use 12! (12 factorial), which is equal to 12 x 11x 10 x9 x8 x7 x6 x5 x4 x3 x2 x1. 12! = 479001600.Then count the of each letter and calculate how many permutations of each letter can be made. For example, here is 1 h, so there is 1 permutation of 1 h.H 1I 1P 60 2T 1A 1M 1U 1S 1Multiply these numbers together. 1 x1 x6 x2 x1 x1 x1 x1 x1 = 12Divide 12! by this number. 479001600 / 12 = 39,916,800 Distinguishable Permutations.
What is the answer to x5?
x5 is an expression: there can be no answer to it.
