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Answer:
The 99% confidence interval estimate of the mean body temperature of all healthy humans is (97.955F, 98.245F).
98.6F is above the upper end of the interval, which means that the sample suggests that the mean body temperature could be lower than 98.6F.
Step-by-step explanation:
The first step is finding the confidence interval
The sample size is 103.
The first step to solve this problem is finding how many degrees of freedom there are, that is, the sample size subtracted by 1. So
Then, we need to subtract one by the confidence level and divide by 2. So:
Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 102 and 0.005 in the t-distribution table, we have .
Now, we need to find the standard deviation of the sample. That is:
Now, we multiply T and s
For the lower end of the interval, we subtract the mean by M. So 98.1 - 0.145 = 97.955F.
For the upper end of the interval, we add the mean to M. So 98.1 + 0.145 = 98.245F.
The 99% confidence interval estimate of the mean body temperature of all healthy humans is (97.955F, 98.245F).
98.6F is above the upper end of the interval, which means that the sample suggests that the mean body temperature could be lower than 98.6F.
So... let's say the amounts invested were "a" at 6% and "b" at 7.5%.
ok.. hmm what's 6% of a? well, (6/100) * a or 0.06a.
what's 7.5% of b? well, (7.5/100) * b or 0.075b.
now... we know, whatever "a" and "b" are, they total the investment of 5000 bucks, thus a + b = 5000
and the interest yielded was 337.50, thus 0.06a + 0.075b = 337.50
thus
solve for "a", to see how much was invested at 6%.
what about "b"? well, b = 5000 - a.
ok.. hmm what's 6% of a? well, (6/100) * a or 0.06a.
what's 7.5% of b? well, (7.5/100) * b or 0.075b.
now... we know, whatever "a" and "b" are, they total the investment of 5000 bucks, thus a + b = 5000
and the interest yielded was 337.50, thus 0.06a + 0.075b = 337.50
thus
solve for "a", to see how much was invested at 6%.
what about "b"? well, b = 5000 - a.