The standard form of a circle is
where h and k are the center and x and y are the coordinates you're given. We need to solve for the radius to finish this off correctly. Filling in, we have
and
, giving us that
. Therefore, our equation is
Answer:
c. The sampling distribution of the sample means can be assumed to be approximately normal because the distribution of the sample data is not skewed
Step-by-step explanation:
From the given data, we have;
The category of the sample = Retired individuals
The number of participants in the sample = 20
The duration of program = six-weeks
The improvement seen by most participants = Little to no improvement
The improvement seen by few participants = Drastic improvement
Therefore, given that the participants are randomly selected and the majority of the participants make the same observation of improvement in the time to walk a mile, we have that, the majority of the outcomes show little difference in walk times after the program, therefore, the distribution of the sample data is not skewed and can be assumed to be approximately normal
Answer:
f(x) = 4.35 +3.95·sin(πx/12)
Step-by-step explanation:
For problems of this sort, a sine function is used that is of the form ...
f(x) = A + Bsin(2πx/P)
where A is the average or middle value of the oscillation, B is the one-sided amplitude, P is the period in the same units as x.
It is rare that a tide function has a period (P) of 24 hours, but we'll use that value since the problem statement requires it. The value of A is the middle value of the oscillation, 4.35 ft in this problem. The value of B is the amplitude, given as 8.3 ft -4.35 ft = 3.95 ft. Putting these values into the form gives ...
f(x) = 4.35 + 3.95·sin(2πx/24)
The argument of the sine function can be simplified to πx/12, as in the Answer, above.
Yes, we apply the distributive property to simplify this expression.
3*5 = 15
6*3 = 18
15 + 18 is the final answer.
