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3 years ago

Y=+4=3/4 (x-1) Find a point on the line and the lines slope

Mathematics

1 answer:

snow_tiger [21]3 years ago

3 0

Answer:

slope = -3/4

Step-by-step explanation:

Slope:

− 3/4

y-intercept:

13 /4

Any line can be graphed using two points. Select two values, and plug them into the equation to find the corresponding values.

x | y

0 | 13 /4

1 | 5/ 2

Hope this helps @(^_^)@

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3 years ago

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What is the midpoint of the line segment with endpoints (-5.5, -6.1) and (-0.5,9.1)?

umka21 [38]

(-3, 1.5)....simply add the x coordinates together, divide them by two, and then do the same for the y coordinates.

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4 years ago

A manufacturer finds that the revenue generated by selling of cortan commodity is given by function R(x)=80x-0.2x^ 2 , where he

Nina [5.8K]

The maximum units is 200 and , Total revenue is $8,000

Step-by-step explanation:

Here we have , A manufacturer finds that the revenue generated by selling of cortan commodity is given by function R(x)=80x-0.2x^ 2 , where he maximum reveremany should be manufactured to obtain this maximum units .Let's find out:

We have following function as $R(x)=80x-0.2x^ 2$ . Let's differentiate this and equate it to zero to find value of x for which the function is maximum!

⇒ $R(x)=80x-0.2x^ 2$

⇒ $\frac{d(R(x))}{dx}=\frac{d(80x-0.2x^ 2)}{dx}$

⇒ $\frac{d(R(x))}{dx}=\frac{d(80x)}{dx}-\frac{d(0.2x^ 2)}{dx}}$

⇒ $0=80-2x(0.2)$

⇒ $\frac{80}{0.4}=x$

⇒ $x=200$

Now , Value of function at x=200 is :

⇒ $R(200)=80(200)-0.2(200)^ 2$

⇒ $R(200)=16000-8000$

⇒ $R(200)=8000$

Therefore , The maximum units is 200 and , Total revenue is $8,000

7 0

3 years ago

Help me please and thank you!

mamaluj [8]

7^6 is the correct answer

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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.

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3 years ago