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AVprozaik [17]
3 years ago
11

Over the last six months, the following numbers of absences have been reported: 10, 16, 12, 18, 2, and 14. What is the median nu

mber of monthly absences?
Mathematics
1 answer:
nasty-shy [4]3 years ago
5 0

Using it's concept, it is found that the median number of monthly absences is of 13.

The <em>median </em>of a data-set is the <u>value that separates the bottom 50% of the data-set from the upper 50%</u>.

The first step is ordering the data-set, hence:

2, 10, 12, 14, 16, 18

In this problem, the cardinality is of 6, which is a even number.

  • Hence, the median is the <u>mean of the 6/2 = 3rd and 4th elements</u>.

Then:

\frac{12 + 14}{2} = 13

The median number of monthly absences is of 13.

To learn more about the median, you can take a look at brainly.com/question/10322579

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3 years ago
The express the area of the entire rectangle
Sonja [21]

Answer:

area= x^2+8x+7

Step-by-step explanation:

a=l*w

l=x+7

w=x+1

a=(x+7)(x+1)

a= x^2+8x+7

5 0
3 years ago
Which angles are coterminal with 7pi/8?<br><br> Select each correct answer
Oksana_A [137]

Answer:

Step-by-step explanation:

-9pi/8

23pi/8

-25pi/8

I think its.

6 0
3 years ago
If _____, use threshold braking any time you need to brake hard.
Inessa [10]

Answer:

The correct option is;

A. You don't have ABS

Step-by-step explanation:

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3 0
3 years ago
What is the quotient 2y^-6y-20/4y+12 ÷ y+5y+6/3y^2+28y+27​
malfutka [58]

Answer with explanation:

 \rightarrow \frac{\frac{2y^2-6 y-20}{4 y+12}}{\frac{y^2+5 y+6}{3 y^2+28 y+27}}\\\\\rightarrow \frac{\frac{y^2-3y-10}{2 y+6}}{\frac{(y+2)(y+3)}{3 y^2+28 y+27}}\\\\\rightarrow \frac{\frac{(y-5)(y+2)}{2 (y+3)}}{\frac{(y+2)(y+3)}{3 y^2+28 y+27}}\\\\\rightarrow \frac{(y-5)(y+2)}{2 (y+3)}} \times {\frac{3 y^2+28 y+27}{(y+2)(y+3)}}\\\\ \rightarrow\frac{(y-5)\times(3 y^2+28 y+27)}{2 (y+3)^2}}

→y²+5y+6

=y²+3 y+2 y+6

=y×(y+3)+2×(y+3)

=(y+2)(y+3)

→y² -3 y-10

=y² -5 y+2 y -10

=y×(y-5)+2×(y-5)

=(y+2)(y-5)

6 0
3 years ago
Read 2 more answers
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