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BARSIC [14]
3 years ago
10

The figure below shows the dimension of an arrow pointed on the pavement of a parking lot. The entire length of the arrow is 8.5

feet. What is the area of the arrow?
A. 20 sq ft.
B. 21 sq ft.
C. 26.75 sq ft
D. 25.5 sq ft
Mathematics
1 answer:
Elodia [21]3 years ago
8 0
Letter b is the right answer choice
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Help please :(??? I need an answer
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C. $23 per day.

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Hope this helps!

8 0
3 years ago
Find the​ (a) mean,​ (b) median,​ (c) mode, and​ (d) midrange for the data and then​ (e) answer the given question. Listed below
mafiozo [28]

Answer:

a) \bar X = 369.62

b) Median=175

c) Mode =450

With a frequency of 4

d) MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5

<u>e)</u>s = 621.76

And we can find the limits without any outliers using two deviations from the mean and we got:

\bar X+2\sigma = 369.62 +2*621.76 = 1361

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

Step-by-step explanation:

We have the following data set given:

49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000

Part a

The mean can be calculated with this formula:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

Replacing we got:

\bar X = 369.62

Part b

Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:

Median=175

Part c

The mode is the most repeated value in the sample and for this case is:

Mode =450

With a frequency of 4

Part d

The midrange for this case is defined as:

MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5

Part e

For this case we can calculate the deviation given by:

s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And replacing we got:

s = 621.76

And we can find the limits without any outliers using two deviations from the mean and we got:

\bar X+2\sigma = 369.62 +2*621.76 = 1361

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

5 0
3 years ago
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