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inessss [21]
3 years ago
13

Consider the graph below:

Mathematics
1 answer:
ladessa [460]3 years ago
4 0

Answer:

y=-5/4x+5

Step-by-step explanation:

The line goes down 5 for every 4 to the right, and it crosses the y-axis at (0,5).

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Suppose on the first day, 80 adult, 120 student, and 20 senior tickets were sold with a total of $1,460 in ticket sales. On the
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She spend $1,238 a month

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Ms. Robinson gave her class 12 minutes to read. Carrie read 5 ½ pages in that time. At what rate, in pages per hour, did Carrie
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3 years ago
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A rectangular yard measuring 39 ft by 67 ft is bordered (and surrounded) by a fence. Inside, a walk that is 4 ft wide goes all t
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The area of the walk is 784 square feet

<h3>What is an area?</h3>

Area is the quantity that expresses the extent of a region on the plane or on a curved surface

First, calculate the area of the rectangular yard.

Area = L x W    

where A equals area, L equals length, and W equals width.

According to the stated problem:

A = 39 x 67

A = 2613 square feet          <-------Area of Rectangular yard.

Now, according to the stated problem there is a walk that is 4 feet wide going all the way along the fence (Inside). "

Therefore, we have 8 feet to subtract from each of the values above (39) and (67).

That is 4 feet on each side,

A = L x W

A = 31 x 59

A = 1829 square feet     <----------Area of new rectangle formed by allowing for three foot walk.

Now, subtract the area of the new rectangle formed from the area of the original area. Doing so, will give you the area of the walk, correct?

Original area    =  2613  square feet

New area         =   1829 square feet

Hence, the Area of WALK = 784 square feet

Learn more about  area on: brainly.com/question/22972014

#SPJ1

3 0
1 year ago
Because of staffing decisions, managers of the Gibson-Marion Hotel are interested in the variability in the number of rooms occu
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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

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