5. Let us apply the formula to some numbers. The following are the durations in minutes of a series of telephone calls. You want
1 answer:
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We have that
2sin²<span>x - sinx - 3 = 0
</span>
Let
A------> sin x
so
2A²-A-3=0
using a graph tool-----> to resolve the second order equation
see the attached figure
the solutions are
A=-1
A=1.5------> is not solution because sin x <span>can not be 1.5
the solution is
A=-1
therefore
sin x=-1
x=arc sin(-1)=-90</span>°
the answer is
sin x=-1
x=-90° or 270°
2sin²<span>x - sinx - 3 = 0
</span>
Let
A------> sin x
so
2A²-A-3=0
using a graph tool-----> to resolve the second order equation
see the attached figure
the solutions are
A=-1
A=1.5------> is not solution because sin x <span>can not be 1.5
the solution is
A=-1
therefore
sin x=-1
x=arc sin(-1)=-90</span>°
the answer is
sin x=-1
x=-90° or 270°
Answer:
Step-by-step explanation:
Multiply the fraction by the conjugate and solve (would explain more but you're on a time crunch)
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.