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2 years ago

Help please need answers

Mathematics

1 answer:

Rzqust [24]2 years ago

4 0

Answer:

slope: 2/3 or 0.6666...

Step-by-step explanation:

find two points on the line : (1,2) and (0,-2) and do rise over run. Rise = 2 and Run = 3, so you put it in rise over run form. Rise/Run = 2/3. The graph is positive.

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Because of staffing decisions, managers of the Gibson-Marion Hotel are interested in the variability in the number of rooms occu

olchik [2.2K]

Answer:

a) $s^2 =30^2 =900$

b) $\frac{(19)(30)^2}{30.144} \leq \sigma^2 \leq \frac{(19)(30)^2}{10.117}$

$567.28 \leq \sigma^2 \leq 1690.224$

c) $23.818 \leq \sigma \leq 41.112$

Step-by-step explanation:

Assuming the following question: Because of staffing decisions, managers of the Gibson-Marimont Hotel are interested in the variability in the number of rooms occupied per day during a particular season of the year. A sample of 20 days of operation shows a sample mean of 290 rooms occupied per day and a sample standard deviation of 30 rooms

Part a

For this case the best point of estimate for the population variance would be:

$s^2 =30^2 =900$

Part b

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The degrees of freedom are given by:

$df=n-1=20-1=19$

Since the Confidence is 0.90 or 90%, the significance $\alpha=0.1$ and $\alpha/2 =0.05$ , the critical values for this case are:

$\chi^2_{\alpha/2}=30.144$

$\chi^2_{1- \alpha/2}=10.117$

And replacing into the formula for the interval we got:

$\frac{(19)(30)^2}{30.144} \leq \sigma^2 \leq \frac{(19)(30)^2}{10.117}$

$567.28 \leq \sigma^2 \leq 1690.224$

Part c

Now we just take square root on both sides of the interval and we got:

$23.818 \leq \sigma \leq 41.112$

5 0

3 years ago

One year Perry had the lowest ERA (earned-run average, mean number of runs yielded per nine innings pitched) of any male pitch

Blizzard [7]

Answer:

z score Perry $z=-1.402$

z score Alice $z=-1.722$

Alice had better year in comparison with Perry.

Step-by-step explanation:

Consider the provided information.

One year Perry had the lowest ERA of any male pitcher at his school, with an ERA of 3.02. For the males, the mean ERA was 4.206 and the standard deviation was 0.846.

To find z score use the formula.

$z=\frac{x-\mu}{\sigma}$

Here μ=4.206 and σ=0.846

$z=\frac{3.02-4.206}{0.846}$

$z=\frac{-1.186}{0.846}$

$z=-1.402$

Alice had the lowest ERA of any female pitcher at the school with an ERA of 3.16. For the females, the mean ERA was 4.519 and the standard deviation was 0.789.

Find the z score

where μ=4.519 and σ=0.789

$z=\frac{3.16-4.519 }{0.789}$

$z=\frac{-1.359}{0.789}$

$z=-1.722$

The Perry had an ERA with a z-score is –1.402. The Alice had an ERA with a z-score is –1.722.

It is clear that the z-score value for Perry is greater than the z-score value for Alice. This indicates that Alice had better year in comparison with Perry.

7 0

2 years ago

*Asymptotes*<br> g(x) =2x+1/x-3 <br><br> Give the domain and x and y intercepts

Nataly [62]

Answer: Assuming the function is $g(x)=\frac{2x+1}{x-3}$ :

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The horizontal asymptote is $y=2$ .

The vertical asymptote is $x=3$ .

Step-by-step explanation:

I'm going to assume the function is: $g(x)=\frac{2x+1}{x-3}$ and not $g(x)=2x+\frac{1}{x}-3$ .

So we are looking at $g(x)=\frac{2x+1}{x-3}$ .

The x-intercept is when y is 0 (when g(x) is 0).

Replace g(x) with 0.

$0=\frac{2x+1}{x-3}$

A fraction is only 0 when it's numerator is 0. You are really just solving:

$0=2x+1$

Subtract 1 on both sides:

$-1=2x$

Divide both sides by 2:

$\frac{-1}{2}=x$

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is when x is 0.

Replace x with 0.

$g(0)=\frac{2(0)+1}{0-3}$

$y=\frac{2(0)+1}{0-3}$

$y=\frac{0+1}{-3}$

$y=\frac{1}{-3}$

$y=-\frac{1}{3}$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The vertical asymptote is when the denominator is 0 without making the top 0 also.

So the deliminator is 0 when x-3=0.

Solve x-3=0.

Add 3 on both sides:

x=3

Plugging 3 into the top gives 2(3)+1=6+1=7.

So we have a vertical asymptote at x=3.

Now let's look at the horizontal asymptote.

I could tell you if the degrees match that the horizontal asymptote is just the leading coefficient of the top over the leading coefficient of the bottom which means are horizontal asymptote is $y=\frac{2}{1}$ . After simplifying you could just say the horizontal asymptote is $y=2$ .

Or!

I could do some division to make it more clear. The way I'm going to do this certain division is rewriting the top in terms of (x-3).

$y=\frac{2x+1}{x-3}=\frac{2(x-3)+7}{x-3}=\frac{2(x-3)}{x-3}+\frac{7}{x-3}$

$y=2+\frac{7}{x-3}$

So you can think it like this what value will y never be here.

7/(x-3) will never be 0 because 7 will never be 0.

So y will never be 2+0=2.

The horizontal asymptote is y=2.

(Disclaimer: There are some functions that will cross over their horizontal asymptote early on.)

6 0

3 years ago

A business has $15,000 to spend on airline tickets to travel to a conference. It wants 27 of its employees to attend. The busine

VikaD [51]

In order to get as many tickets they need 15 business class tickets and 12 economy class tickets (700x15=10,500)(375x12=4500) Which when you add those two makes 15,000.

4 0

3 years ago

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What is the value of 4 to the 3 power divided by 2 to the third power?

Sholpan [36]

Answer:

Please Vote Me Brainliest. Thank You.

7 0

1 year ago

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Help please need answers ​

Help please need answers