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sergey [27]
2 years ago
7

The box plots below show student grades on the most recent exam compared to overall grades in a class:

Mathematics
1 answer:
777dan777 [17]2 years ago
3 0

The statement that best describes the information about the medians would be: the class median and exam median are almost the same.

<h3>Median of a Data Set on a Box Plot</h3>
  • The median of any data distribution that is plotted on a box plot is indicated by the vertical line that divides the rectangular box in the box plot.

Thus:

  • The median for class is 84
  • The median for exam is 85.

Therefore, the statement that best describes the information about the medians would be: the class median and exam median are almost the same.

Learn more about Box Plot on:

brainly.com/question/10209877

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I need help please help
Mademuasel [1]

Answer:

Step-by-step explanation:

Note that 28 + 110 + 42 = 180, and that 28 + 42 + 110 = 180 also.

Since all three angles of one triangle are the same as the corresponding angles of the other triangle, the triangles are similar.

8 0
3 years ago
The citizens of a certain community were asked to choose their favorite pet. The pie chart below shows the distribution of the c
Arturiano [62]

Incomplete Question:

The content of the pie chart is as follows:

<em>Hamsters  = 9% ; Snakes  = 10% ; Cats  = 23% </em>

<em> Birds  = 21% ;  Dogs  = 26% ;  Fish  = 11%</em>

<em></em>

Answer:

The number of citizens who chose cat or fish is 47,600

Step-by-step explanation:

Given

Number of citizens = 140,000

Required

Determine the number of those that chose fish or cats

First, we need to calculate the percentage of those whose pets are either cats or fish

Percentage = Cat + Fish

Substitute 23% for cat and 11% for fish

Percentage = 23\% + 11\%

Percentage = 34\%

Next, is to multiply the calculated percentage by the number of citizens

Cat\ or\ Fish = Percentage * Number\ of\ Citizens

Cat\ or\ Fish = 34\% * 140000

Cat\ or\ fish = 47600

<em>Hence, the number of citizens who chose cat or fish is 47,600</em>

4 0
3 years ago
Give one example of when you would make a frequency table to solve the problem
RideAnS [48]

Answer:

Consider a class in which there are 40 students .There is a Mathematics test of 50 marks .Two students were absent.The Duration of test was for one and half an hour.

So Marks of students according to their roll number are 12,8,4,20,25,36,2,29,35,39,5,48,43,37,38,29,17,15,15,25,31,32,26,28,27,24,2,15,34,,50,46,48,37,28,26,45,49,24.

So in this case it would be better to prepare frequency polygon by selecting a class interval.

Class interval = 0-5,5-10,10-15,15-20,20-25,25-30,30-35,35-40,40-45,45-50.

Or 0-10,10-20,20-30,30-40,40-50.

or you can make class interval by yourself.





4 0
3 years ago
4x-1=y graphed by using a table
olasank [31]

Answer:

It can be many answers since it’s on a graph

Step-by-step explanation:

That equation isn’t the rule so test it out on number so and you will get point to put on a graph

5 0
3 years ago
Quadrilateral ABCD is a rectangle. If AG = -7j + 7 and DG = 5j + 43, find BD
Kaylis [27]

<em>BD</em> = 56

Step-by-step explanation:

Step 1: In rectangle, the diagonals are congruent and bisect each other.

So, <em>AC</em> = <em>BD</em>

⇒<em>AG</em> + <em>GC</em> = <em>BG</em> + <em>GD</em>

⇒<em>AG</em> + <em>AG</em> = <em>GD</em> + <em>GD</em>

⇒ 2<em>AG</em> = 2<em>GD</em>

⇒<em>AG</em> = <em>GD</em>

⇒ –7<em>j </em>+ 7 = 5<em>j</em> + 43

⇒–7<em>j</em> – 5<em>j</em> = 43 – 7

⇒–12<em>j</em> = 36

⇒<em>j</em> = –3

Step 2: <em>BD</em> = 2<em>DG</em>

<em>BD</em> = 2(5<em>j</em> + 43)

     = 2(5 (–3) + 43)

     = 2(–15 + 43)

     = 2 × 28

     = 56

Hence, <em>BD</em> = 56.

4 0
3 years ago
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