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3 years ago

If angle a and angle b are complementary angles, they they share a common side.

Mathematics

2 answers:

asambeis [7]3 years ago

5 0

Answer:

However it is not necessary that if they are complementary angles they must share a side.Any angles that add up to 90 degree,can be considered complementary.

Snezhnost [94]3 years ago

3 0

Answer:

False.

Step-by-step explanation:

A complementary angle is an angle where two seperate angles add together for a 90 degree angle. They usually share a side, but not always.

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The accompanying stem-and-leaf plot represents Kiko's test scores this year. What is the median score for this set of data?

worty [1.4K]

Answer: 80

Write down all the values and cross out until you get to the middle, since there are 2 values, find the mean, which would be 80

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4 years ago

A 16−oz bottle of parmesan cheese costs $5.78, while an 8−oz bottle costs $3.97. Which is the better buy? Round your answer to t

LUCKY_DIMON [66]

Answer:

The 16-oz bottle of parmesan cheese is the better buy.

Step-by-step explanation:

Even though 8-oz is half the amount of 16-oz, that doesn't mean it's cheaper. Let's find the amount of and 8-oz bottle when a 16-oz bottle is $5.78:

16-oz bottle of parmesan cheese = $5.78

A price of an 8-oz bottle when 16-oz is divided by 2:

16 ÷ 5.78 = $2.76

The original price of the 8-oz bottle should've been $2.76 but it's more expensive due to VAT.

So the 8-oz bottle is more expensive, therefore the 16-oz bottle is the better buy!

8 0

3 years ago

A sample of 200 college freshmen was asked how many hours per week they spent playing video games. The following frequency distr

JulsSmile [24]

Answer:

$\% Below\ 50 = 88 \%$

$\% 43\ or\ more = 17 \%$

Step-by-step explanation:

Given

$Total = 200$

<em>The attachment presents the table properly</em>

Solving (a): Percentage of those who play for less than 50 hours weekly

To do this, we simply consider the frequency of students that fall within:

1.0 - 7.9, 8.0 - 14.9,......,43.0 - 49.9

Their frequency is:

$Frequency = 21 + 37 + 40 + 27 + 23 + 18 + 10$

$Frequency = 176$

The percentage is then calculated as:

$\% Below\ 50 = \frac{Frequency}{Total} * 100\%$

$\% Below\ 50 = \frac{176}{200} * 100\%$

$\% Below\ 50 = \frac{176* 100}{200} \%$

$\% Below\ 50 = \frac{176* 1}{2} \%$

$\% Below\ 50 = \frac{176}{2} \%$

$\% Below\ 50 = 88 \%$

Solving (b): Percentage of those who play for 43 or more hours weekly

To do this, we simply consider the frequency of students that fall within:

43.0 - 49.9,......,64.0 - 70.9

Their frequency is:

$Frequency = 10 + 7 + 5+12$

$Frequency = 34$

The percentage is then calculated as:

$\% 43\ or\ more = \frac{Frequency}{Total} * 100\%$

$\% 43\ or\ more = \frac{34}{200} * 100\%$

$\% 43\ or\ more = \frac{34* 100}{200} \%$

$\% 43\ or\ more = \frac{34* 1}{2} \%$

$\% 43\ or\ more = \frac{34}{2} \%$

$\% 43\ or\ more = 17 \%$

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3 years ago

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Answer:

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Answer:

D.27

Step-by-step explanation:

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