Answer:
Here is a graph of that equation.
Let me know if you need more help! :)
Answer:
There is enough evidence to say that the true average heat output of persons with the syndrmoe differs from the true average heat output of non-sufferers.
Step-by-step explanation:
We have to perform a hypothesis test on the difference between means.
The null and alternative hypothesis are:

μ1: mean heat output for subjects with the syndrome.
μ2: mean heat output for non-sufferers.
We will use a significance level of 0.05.
The difference between sample means is:

The standard error is

The t-statistic is

The degrees of freedom are

The critical value for a left tailed test at a significance level of 0.05 and 16 degrees of freedom is t=-1.746.
The t-statistic is below the critical value, so it lies in the rejection region.
The null hypothesis is rejected.
There is enough evidence to say that the true average heat output of persons with the syndrmoe differs from the true average heat output of non-sufferers.
Answer:
I’m doing that exact question right now on a test so I just guessed
Step-by-step explanation:
Explanation:
For the purpose of filling in the table, the BINOMPDF function is more appropriate. The table is asking for p(x)--not p(n≤x), which is what the CDF function gives you.
If you want to use the binomcdf function, the lower and upper limits should probably be the same: 0,0 or 1,1 or 2,2 and so on up to 5,5.
The binomcdf function on my TI-84 calculator only has the upper limit, so I would need to subtract the previous value to find the table entry for p(x).