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AlekseyPX
3 years ago
5

Solve this equation for x 0.79=log x

Mathematics
1 answer:
Vitek1552 [10]3 years ago
8 0
X=6.16595001.........
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Find the center and radius of the circle with equation (x + 7) + (y + 5)2 = 81.
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3 years ago
Confidence Interval Mistakes and Misunderstandings—Suppose that 500 randomly selected recent graduates of a university were as
kvv77 [185]

Answer:

The correct 95% confidence interval is (8.4, 8.8).

Step-by-step explanation:

The information provided is:

n=500\\\bar x=8.6\\\sigma=2.2

(a)

The (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:

CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}

The 95% confidence interval for the average satisfaction score is computed as:

8.6 ± 1.96 (2.2)

This confidence interval is incorrect.

Because the critical value is multiplied directly by the standard deviation.

The correct interval is:

8.6\pm 1.96 (\frac{2.2}{\sqrt{500}})=8.6\pm 0.20=(8.4,\ 8.6)

(b)

The (1 - <em>α</em>)% confidence interval for the parameter implies that there is (1 - <em>α</em>)% confidence or certainty that the true parameter value is contained in the interval.

The 95% confidence interval for the mean rating, (8.4, 8.8) implies that the true there is a 95% confidence that the true parameter value is contained in this interval.

The mistake is that the student concluded that the sample mean is contained in between the interval. This is incorrect because the population is predicted to be contained in the interval.

(c)

The (1 - <em>α</em>)% confidence interval for population parameter implies that there is a (1 - <em>α</em>) probability that the true value of the parameter is included in the interval.

The 95% confidence interval for the mean rating, (8.4, 8.8) implies that the true mean satisfaction score is contained between 8.4 and 8.8 with probability 0.95 or 95%.

Thus, the students is not making any misinterpretation.

(d)

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

In this case the sample size is,

<em>n </em>= 500 > 30

Thus, a Normal distribution can be applied to approximate the distribution of the alumni ratings.

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3 years ago
At the end of the year a car is worth 5% less than what it was worth at the beginning of the year. If a car was worth $9 500 in
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The answer should be 9025
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3 years ago
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Evaluate the expression when d = 20 and v = 19 20v + 12d - 6v
Alexeev081 [22]
D = 20 , v = 19
20v + 12d - 6v
= 20 (19) + 12 (20) - 6 (19)
= 380 + 240 - 114
= 620 - 114
= 506
When d = 20 and v = 19
20v + 12d - 6v = 506
6 0
2 years ago
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What element is 1s22s22p63s23p63d104s24p65s24d105p6
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Answer:

carbon

Step-by-step explanation:

it is obvious you can tell

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