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liberstina [14]
4 years ago
15

What is the value of 7x-2x+3,when x=2

Mathematics
2 answers:
borishaifa [10]4 years ago
8 0
Well if x = 2, then you do 7 times 2, which is 14, then 2 times 2 which is 4, then 14 minus 4 is 10, and 10 plus 3 is 13.
ollegr [7]4 years ago
5 0
BIDMAS

2x = 4 
4 + 3 = 7
7x = 14
14 - 7 = 7
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Why is the inequality x≤6 equivalent to the inequality x−5≤6−5
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Two inequalities are equivalent if they have the same set of solutions.

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x ≤ 6

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Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true
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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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