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dusya [7]
3 years ago
12

PLEASE HELP ASAP

Mathematics
1 answer:
NeX [460]3 years ago
4 0

Answer:

(a) Equation of line n is y=\frac{1}{2}x. (b)  Equation of line p is y=\frac{1}{2}x-5. (c)  Equation of line r is y=x.

Explanation:

The equation of line m is

y=\frac{1}{2}x-5

It is a slope intercept form of a line, therefore the slope of the line is \frac{1}{2}.

(a)

Two line have no solution if and only if both lines are parallel. The slope of two parallel lines are same, therefore the slope of the line n is must be \frac{1}{2}.

The equation for line n is in the form of

y=\frac{1}{2}x+c

Where c can be any real number except -5.

An equation for line n is

y=\frac{1}{2}x

(b)

Two line have infinitely many solutions if and only if both lines are same.

Therefore the equation of line p is same as equation of line m.

y=\frac{1}{2}x-5

(c)

Two line have exactly one solution if and only if both lines intersecting each other at a single point. To lines intersect each other if their slopes are different.

Therefore the slope of line r is not equal to \frac{1}{2}.

The equation for line r is in the form of

y=mx+c

Where c can be any real number and m can be any real number except \frac{1}{2}.

An equation for line r is

y=x

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Can the sum of two irrational numbers ever be a rational number?
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4 years ago
Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true
Feliz [49]

Answer:

a) The 95% CI for the true average porosity is (4.51, 5.19).

b) The 98% CI for true average porosity is (4.11, 5.01)

c) A sample size of 15 is needed.

d) A sample size of 101 is needed.

Step-by-step explanation:

a. Compute a 95% CI for the true average porosity of a certain seam if the average porosity for 20 specimens from the seam was 4.85.

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

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

M = 1.96*\frac{0.78}{\sqrt{20}} = 0.34

The lower end of the interval is the sample mean subtracted by M. So it is 4.85 - 0.34 = 4.51

The upper end of the interval is the sample mean added to M. So it is 4.35 + 0.34 = 5.19

The 95% CI for the true average porosity is (4.51, 5.19).

b. Compute a 98% CI for true average porosity of another seam based on 16 specimens with a sample average of 4.56.

Following the same logic as a.

98% C.I., so z = 2.327

M = 2.327*\frac{0.78}{\sqrt{16}} = 0.45

4.56 - 0.45 = 4.11

4.56 + 0.45 = 5.01

The 98% CI for true average porosity is (4.11, 5.01)

c. How large a sample size is necessary if the width of the 95% interval is to be 0.40?

A sample size of n is needed.

n is found when M = 0.4.

95% C.I., so Z = 1.96.

M = z*\frac{\sigma}{\sqrt{n}}

0.4 = 1.96*\frac{0.78}{\sqrt{n}}

0.4\sqrt{n} = 1.96*0.78

\sqrt{n} = \frac{1.96*0.78}{0.4}

(\sqrt{n})^{2} = (\frac{1.96*0.78}{0.4})^{2}

n = 14.6

Rounding up

A sample size of 15 is needed.

d. What sample size is necessary to estimate the true average porosity to within 0.2 with 99% confidence?

99% C.I., so z = 2.575

n when M = 0.2.

M = z*\frac{\sigma}{\sqrt{n}}

0.2 = 2.575*\frac{0.78}{\sqrt{n}}

0.2\sqrt{n} = 2.575*0.78

\sqrt{n} = \frac{2.575*0.78}{0.2}

(\sqrt{n})^{2} = (\frac{2.575*0.78}{0.2})^{2}

n = 100.85

Rounding up

A sample size of 101 is needed.

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4 years ago
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