Given the assumption of a normal distribution with a mean of 100 and a standard deviation of 15, a score of 130 represents a z-score of 2. In different mostly older handbooks of descriptive statistics you can find a ‘from z to percentile-table’. Nowadays it’s more easy to use the computer. Using R, a free and open source statistical environment (www.r-project.org), you can use the command pnorm (130, 100, 15) and it will give 0.9772. Because yo want the proportion above that score you use 1-pnorm (130, 100, 15). Another way of writing in R and with for example 3 IQ-scores:
perc = pnorm (c (70, 100, 130), 100, 15)
(1 - perc)
gives you the above-proportion of the IQ-scores of respectively 70, 100 and 130.
I’m guessing it’s going to be third power
N = number of compounding periods
Years = log (total / principal) / n * log (1 + rate / n)
Years = log (750 / 500) / 4 * log (1 + .025/n)
Years = log (1.5) / 4 * log (1<span><span>.00625)
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</span> <span>Years = 0.17609125906 / 4 * 0.0027058933759
</span><span>Years = 0.17609125906
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0.0108235735
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Years =
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16.2692348382
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Source Calculator
http://www.1728.org/compint.htm
Answer:The answer of 2 into 2 is 4
Step-by-step explanation:
2×2=4