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ExtremeBDS [4]
2 years ago
15

A group of students were surveyed to find out if they like building snowmen or skiing as a winter activity. The results of the s

urvey are shown below:
60 students like building snowmen
10 students like building snowmen but do not like skiing
80 students like skiing
50 students do not like building snowmen

Make a two-way table to represent the data and use the table to answer the following questions.

Part A: What percentage of the total students surveyed like both building snowmen and skiing? Show your work. (5 points)
Mathematics
2 answers:
soldier1979 [14.2K]2 years ago
8 0
Building Snowmen - 60
Building Snowmen (no skiing) - 10
Skiing - 80
No to building snowmen - 50

Part A:
total of students: 200
140 x
----- = ----
200 100
14000 200x
--------- = --------
200 200
70 = x
70%
adelina 88 [10]2 years ago
7 0

Answer:

The proportion of students who like both building snowmen and skiing

50/110 

percent is 50/110 * 100 = 45%

part b) 

Of the students who don't like building snowmen,

what is the probability of choosing a student who does not like skiing as well?

There are 50 students who don't like building snowmen. 

There are 20 students who don't like skiing and don't like building snowmen.

probability is is 20/50 = 2/5

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Hope I helped :) 


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Step-by-step explanation:

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3 years ago
Verify that (cos²a) (2 + tan² a) = 2 - sin² a....<br>​
Sveta_85 [38]

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\boxed{\sf \ \sf \sin^2 \theta + \cos^2 \theta = 1}

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Given to verify the following:

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\texttt{\underline{apply distributive method}:}

\rightarrow \sf 2 (cos^2a) + (\dfrac{sin^2 a}{cos^2 a} ) (cos^2a)

\texttt{\underline{simplify the following}:}

\rightarrow \sf 2cos^2 a  + sin^2 a

\texttt{\underline{rewrite the equation}:}

\rightarrow \sf 2(1 - sin^2a )  + sin^2 a

\texttt{\underline{distribute inside the parenthesis}:}

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Only smart bros know the answer
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33.47

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