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babymother [125]
4 years ago
11

What is the area of the composite figure?

Mathematics
2 answers:
Arte-miy333 [17]4 years ago
6 0
34 units got it right on edgeunity
Irina18 [472]4 years ago
4 0

Answer:

area of the compostie figure =  34 squre units

Step-by-step explanation:

To find the area of this composite figure , first find the area of the trapezoid

then we subtract the area of rectangle (white part of the figure at the bottom)

Area of the trapezoid = (base1+base2)/2  times height

base 1= 10

base 2=6, height = 5

Area of a trapezoid = \frac{10+6}{2} *5= 40

Area of the rectangle = length * width

= 3*2= 6

Now area of the compostie figure = 40 - 6= 34 squre units

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tino4ka555 [31]

Given:

The two ways table.

To find:

The number of girls in year 9.

Solution:

Using the two ways table, we get

Boys in year 10 = Total student in year 10 - Girls in year 10

                        = 256 - 123

                        = 133

Boys in year 9 = Total boys - Boys in year 10 - Boys in year 11

                        = 407 - 133 - 125

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Girls in year 9 = Total students in year 9 - Boys in year 9

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                       = 154

Therefore, the number of girls in year 9 is 154 and the correct option is C.

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3 years ago
Why can you use a model to check your answer for a division problem
dangina [55]
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8 0
3 years ago
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mamaluj [8]

Answer:

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Step-by-step explanation:

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8 0
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Read 2 more answers
Probability extension: Jerry did only 5 problems of one assignment. What is the probability that the problems he did comprised t
zimovet [89]

Answer:

(b) 1/792

Step-by-step explanation:

The complete question is;

<em>Counting: Grading One professor grades homework by randomly choosing 5 out of 12 homework problems to grade.</em>

<em>(a) How many different groups of 5 problems can be chosen from the 12</em>

<em>problems?</em>

<em>(b)Probability extension: Jerry did only 5 problems of one assignment. What is the probability that the problems he did comprised the group that was selected to be graded?</em>

<u>In (a)</u>

<u></u>

Apply the formula

\frac{n!}{(n-r)!(r!)}

where n=12 and r=5

substitute values

=\frac{12!}{(12-5)!(5!)} \\\\\\=\frac{12!}{(7!)(5!)} \\\\\\=\frac{12*11*10*9*8}{5*4*3*2*1} \\\\\\=\frac{95040}{120} \\\\\\=792

In (b)

If Jerry did only 5 problems of one assignment then the probability  will be

\frac{5}{12} *\frac{4}{11} *\frac{3}{10} *\frac{2}{9} *\frac{1}{8} =\frac{1}{792}

<em />

<em />

4 0
3 years ago
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