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g100num [7]
3 years ago
13

What are the possible solutions to 33⋅3333⋅33, if 11⋅11=411⋅11=4 and 22⋅22=1622⋅22=16?

Mathematics
1 answer:
SOVA2 [1]3 years ago
3 0

Since 11*11 = 4, therefore in this case 11 = 2. Since 22*22 = 16, therefore in this case 22 = 4. So we can see that the common ratio is 2. Based from this, we can say that 33 = 6.

So that:

33*33 = 36

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A sample of 10 adult elephants had an average weight of 12,556 pounds. The standard deviation
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The 99% confidence interval of the population mean for the weights of adult elephants is between 12,475 pounds and 12,637 pounds.

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We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

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The upper end of the interval is the sample mean added to M. So it is 12,556 + 81 = 12,637 pounds.

The 99% confidence interval of the population mean for the weights of adult elephants is between 12,475 pounds and 12,637 pounds.

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