
is continuous over its domain, all real

.
Meanwhile,

is defined for real

.
If

, then we have

as the domain of

.
We know that if

and

are continuous functions, then so is the composite function

.
Both

and

are continuous on their domains (excluding the endpoints in the case of

), which means

is continuous over

.
Answer:
It would be changing by which a person thinks.
But, I will pick 569.124.
It would be 5.69124e2. (or 5.6912 x 10^3)
Answer:
The sample mean is
b.3.55
The margin of error is
0.32
Step-by-step explanation:
Deep explanation about a confidence interval
We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of
.
So it is z with a pvalue of
, so 
Now, find M as such

In which
is the standard deviation of the population and n is the size of the sample.

The lower end of the interval is the mean subtracted by M. So it is 6.4 - 0.3944 = 6.01 hours.
The upper end of the interval is the mean added to M. So it is 6.4 + 0.3944 = 6.74 hours.
In this problem:
The deep explanation is not that important.
We just have to recognize that the interval has a lower end and an upper end. The distance from both the upper and the lower end to the mean is M. This means that the sample mean is the halfway point between the lower end and the upper end.
The margin of error is the distance of these two points(lower and upper end) to the mean.
In our interval
Lower end: 3.23
Upper end: 3.87
Sample mean

So the correct answer is:
b.3.55
The margin of error is
3.87 - 3.55 = 3.55 - 3.23 = 0.32
Answer:
Yes
Step-by-step explanation:greatest common factor (GCF) of 10 and 14 is 2. We will now calculate the prime factors of 10 and 14, then find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 10 and 14.
Area of the rectangular mat = 12 x 4 = 48 in²
Area of the 3 circles = 3( πr²) = 3(3.14 x 2²) = 37.68 in²
P(landing in a circle) = 37.68/48 = 157/200 = 0.79 (nearest hundredth)
Answer: The probability of landing in one of the circles is 0.79.