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Andru [333]
3 years ago
12

The two intervals (114.9, 115.7) and (114.6, 116.0) are confidence intervals (computed using the same sample data) for μ = true

average resonance frequency (in hertz) for all tennis rackets of a certain type. (a) What is the value of the sample mean resonance frequency?
Mathematics
1 answer:
kenny6666 [7]3 years ago
4 0

Answer:

115.3 hertz

Step-by-step explanation:

For either interval, the value of the sample mean is given by the average of the lower and upper bound of the confidence interval. Since both intervals were constructed by using the same sample, both values should be equal.

For the first interval:

M=\frac{114.9 + 115.7}{2} \\M=115.3\ hertz

For the second interval:

M=\frac{114.6 + 116.0}{2} \\M=115.3\ hertz

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Dima020 [189]
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The solution is where the two graphs intersect, which is
(-3,1).

EXPLANATION

The given system of equations are
y = \frac{1}{3} x + 2
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y = - x - 2

We need to graph the two equations.

Let us graph

y = \frac{1}{3} x + 2
first.

We need at least two points.

You can choose any appropriate value for x and solve for y. Choosing zero makes our working easier. So let us plot the intercepts.

When
x = 0

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(0,2)

When
y = 0
we get,

0= \frac{1}{3} x + 2

\Rightarrow \: - 2 = \frac{1}{3} x

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\Rightarrow \: -6= x
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(-6,0).

We plot these two points and draw a straight line through them to obtain the blue graph in the attachment.

For the second line

y = - x - 2

We again find the intercepts and plot them.

When
x = 0

y = - 0 - 2

\Rightarrow \: y = - 2

This gives the ordered pair

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y = 0

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0 = - x - 2

2 = - x

x = - 2

Then we again have the ordered pair,

(-2,0)

We plot these two points on the same graph sheet to obtain the red graph above.

The intersection of the two lines is
(-3,1)


You will get good grades so don't worry much.

4 0
3 years ago
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