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Vsevolod [243]
2 years ago
9

There are 150 students and 3/5 of these students are girls. 2/3 play school activities. how many play activities?

Mathematics
2 answers:
Murrr4er [49]2 years ago
8 0

Answer:

100 or 60

Step-by-step explanation:

Since the question does not specify whether \frac{2}{3} refers to only the girls or all students, both will be calculated for your convenience.

If 2/3 refers to all students:

150*\frac{2}{3}

=100

If 2/3 refers to the girls:

150*\frac{3}{5}*\frac{2}{3}

=150*\frac{2}{5}

=60

torisob [31]2 years ago
7 0

Answer:

60 girls play activities

Step-by-step explanation:

150/30= 3/5 = 90/150

2/3 =60/90

60 girls

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59

Step-by-step explanation:

We know that all the angles must add up to 180

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38+2b-20+b-15=180

solve for b

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3b=177

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Each angle of an equiangular triangle is what degree angle
Alexandra [31]
An equiangular triangle adds up to 180 so divide that by 3, which is the number of sides, and you get, 60, therefore, each side is 60<span>° angle.</span>
5 0
3 years ago
Mr Cole has 30 students in his class. He has twice as many girls than boys. What is the ratio of girla to boys?
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12:1/2(3)

Step-by-step explanation: no problem homie

8 0
2 years ago
A running circuit is in the shape of a triangle with lengths of 6km, 6.5km and 7km. What are the sizes of the angles (in minutes
Rudik [331]

A <u>triangle</u> is an example of a class of <em>figures</em> referred to as <em>plane shapes</em>. It has <u>three</u> straight <u>sides</u> and <u>three</u> internal <u>angles</u> which sum up to 180^{o}. The <em>measures</em> of the internal <u>angles</u> of the <u>triangle</u> given in the question are A = 52.6^{o}, B = 59.4^{o}, and C = 68^{o}.

A <u>triangle</u> is an example of a class of <em>figures</em> referred to as <em>plane shapes</em>. It has <u>three</u> straight <u>sides</u> and <u>three</u> internal <u>angles</u> which sum up to 180^{o}.

Considering the given question, let the <u>sides</u> of the triangle be: a = 6 km, b = 6.5 km, and c = 7 km.

Apply the <em>Cosine rule</em> to have:

c^{2} = a^{2} + b^{2} - 2ab Cos C

So that;

7^{2} = 6^{2} + (6.5)^{2} - 2(6 * 6.5) Cos C

49 = 36 + 42.25 - 78Cos C

78 Cos C = 78.25 - 49

               = 29.25

Cos C = \frac{29.25}{78}

         = 0.375

C = Cos^{-1} 0.375

   = 67.9757

C = 68^{o}

Apply the <em>Sine rule</em> to determine the <u>value</u> of B,

\frac{b}{Sin B} = \frac{c}{Sin C}

\frac{6.5}{Sin B} = \frac{7}{Sin 68}

SIn B = \frac{6.5 *Sin 68}{7}

         = 0.861

B = Sin^{-1} 0.861

   = 59.43

B = 59.4^{o}

Thus to determine the value of A, we have;

A + B + C = 180^{o}

A + 59.4^{o} + 68^{o} = 180^{o}

A = 180^{o} - 127.4

  = 52.6

A = 52.6^{o}

Therefore the <u>sizes</u> of the <em>internal angles</em> of the triangle are: A = 52.6^{o}, B = 59.4^{o}, and C = 68^{o}.

For more clarifications on applications of the Sine and Cosine rules, visit: brainly.com/question/14660814

#SPJ1

8 0
1 year ago
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