Explanation:
In order to prove that affirmation, we define the function g over the interval [0, 1/2] with the formula 
If we evaluate g at the endpoints we have
g(0) = f(1/2)-f(0) = f(1/2) - f(1) (because f(0) = f(1))
g(1/2) = f(1) - f(1/2) = -g(0)
Since g(1/2) = -g(0), we have one chance out of three
- g(0) > 0 and g(1/2) < 0
- g(0) < 0 and g(1/2) > 0
- g(0) = g(1/2) = 0
We will prove that g has a zero on [0,1/2]. If g(0) = 0, then it is trivial. If g(0) ≠ 0, then we are in one of the first two cases, and therefore g(0) * g(1/2) < 0. Since f is continuous, so is g. Bolzano's Theorem assures that there exists c in (0,1/2) such that g(c) = 0. This proves that g has at least one zero on [0,1/2].
Let c be a 0 of g, then we have
Hence, f(c+1/2) = f(c) as we wanted.
5(2x+3) by adding them you get this equation.
Answer:
Container 
will have less label area than container 
by about 
.
Step-by-step explanation:
A rectangular sheet of paper can be rolled into a cylinder. Conversely, the lateral surface of a cylinder can be unrolled into a rectangle- without changing the area of that surface.
Indeed, the width of that rectangle will be the same as the height of the cylinder. On the other hand, the length of that rectangle should be exactly equal to the circumference of the circles on the top and the bottom of the cylinder. In other words, if a cylinder has a height of 
and a radius of 
at the top and the bottom, then its lateral surface can be unrolled into a rectangle of width and length 
.
Apply this reasoning to both cylinder 
and :
For cylinder , 
while 
. Therefore, when the lateral side of this cylinder is unrolled:
- The width of the rectangle will be
, while - The length of the rectangle will be
.
That corresponds to a lateral surface area of 
.
For cylinder , 
while 
. Similarly, when the lateral side of this cylinder is unrolled:
- The width of the rectangle will be
, while - The length of the rectangle will be
.
That corresponds to a lateral surface area of 
.
Therefore, the lateral surface area of cylinder is smaller than that of cylinder by 
.
Answer:
We conclude that the mean nicotine content is less than 31.7 milligrams for this brand of cigarette.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 31.7 milligrams
Sample mean, 
= 28.5 milligrams
Sample size, n = 9
Alpha, α = 0.05
Sample standard deviation, s = 2.8 milligrams
First, we design the null and the alternate hypothesis
We use One-tailed t test to perform this hypothesis.
Formula:
Putting all the values, we have
Now, 
Since,
We fail to accept the null hypothesis and accept the alternate hypothesis. We conclude that the mean nicotine content is less than 31.7 milligrams for this brand of cigarette.
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