How much is 8% of 1,249? Round the answer to the nearest hundredth.

The graph of the equation 40.51x+12.45y=666.64 is attached with the answer where the horizontal axis represents the X axis and the vertical axis represents Y axis.

To plot the graph physically just find two points lying on the line. Mark the points on the graph sheet and then join them. This will give you the line represented by the equation.

To find points on the line assume the value of any one variable, substitute it in the equation, then solve the equation to find the value of other variable. For example : assume y = 1; substitute the value of y in the equation;

⇒ 40.51x + 12.45×1 = 666.64

⇒ 40.51x = 666.64 - 12.45

⇒ 40.51x = 654.19

⇒ x = $\frac{654.19}{40.51}$

⇒ x ≈ 16.149

Therefore point ( 16.149 , 1 ) lie on the graph of the equation.

***Only two points are required to plot this graph just because it represents a straight line, that we can conclude just by observing the equation. If in an equation the power of x is 1 or 0 and power of y is 1 or 0 then only it will represent a straight line in 2-D plane.***

5 0

3 years ago

Confidence Interval Mistakes and Misunderstandingsâ€”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 - α)% confidence interval for population mean (μ) 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 - α)% confidence interval for the parameter implies that there is (1 - α)% 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 - α)% confidence interval for population parameter implies that there is a (1 - α) 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 μ and standard deviation σ and we take appropriately huge random samples (n ≥ 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,

n = 500 > 30

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

7 0

3 years ago