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svet-max [94.6K]
1 year ago
14

I need help with question 2

Mathematics
1 answer:
Volgvan1 year ago
7 0

2. Area of the entire figure = 14f + 20 units².

Perimeter of the entire figure = 6f + 18 units².

3. D. 8.

4. 11/15

<h3>How to Find the Perimeter and Area of a Rectangle?</h3>

Perimeter = 2(length + width)

Area = (length)(width).

2. Decompose the shape into two rectangles, rectangle 1 and rectangle 2.

Length of rectangle 1 = f

Width of rectangle 1 = 4

Area of rectangle 1 = (f)(4) = 4f units²

Length of rectangle 2 = 2f + 4

Width of rectangle 2 = 5

Area of rectangle 2 = (2f + 4)(5) = 10f + 20 units²

Area of the entire figure = 4f + 10f + 20

Area of the entire figure = 14f + 20 units².

Perimeter of the entire figure = 2f + 4 + 5 + f + 4 + f + 2f + 5

Add like terms

Perimeter of the entire figure = 6f + 18 units².

3. A factor of 44 is any number that can divide 44 without a remainder. 8 is not a factor of 44 because it cannot divide 44 without a remainder. So, the answer is: D. 8.

4. 33/45 = (3 × 11) / (3 × 15)

= 11/15

Learn more about perimeter and area on:

brainly.com/question/19819849

#SPJ1

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Answer:

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====================================================

Explanation:

The set of given numbers sorts to this

11, 50, 72, 78, 80, 81, 85, 89, 94, 96, 97, 98

The majority or main cluster of the values are between 50 and 98. The outlier 11 is off on its own pretty far from the main cluster.

Let's find the mean of every value in this set. If you added up all the numbers, then you should get 931. Divide this over the number of values (12) and the mean of this set is roughly 931/12 = 77.5833

Now let's kick out the outlier 11. The 931 will drop to 931-11 = 920. Divide that new smaller sum over the 12-1 = 11 items left to get 920/11 = 83.6364 approximately.

--------------

To summarize so far:

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  • new mean (removed outlier) = 83.6364 approximately

We see that <u>the mean has increased</u>. This is because the outlier was pulling down on the mean to make it smaller than it should be. Once the outlier is out of the picture, the new mean will increase and better represent the center of the main cluster.

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