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3 years ago

What are th minimum,first quartile,median,third quartile,and maximum of the data set? 12,6,8,3,10,15,18,7

Mathematics

2 answers:

Arturiano [62]3 years ago

5 0

Minimum: 3
First Q: 6
Median: 9
Third Q:15
Maximum: 18

masya89 [10]3 years ago

3 0

You have to put it in order first so it would be:
3,6,7,8,10,12,15,18
Minimum: 3
Quartile 1: 6.5
Quartile 2: 9
Quartile 3: 13.5
Maximum: 18

After that, you make a Box Plot and put it on there. I was doing this in class today. :)

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Mese questions are based on a 52

Mrrafil [7]

Answer:1.) 6/52

Step-by-step explanation:

5 0

3 years ago

9-8x=35 explain please

ANEK [815]

9 - 8x = 35

subtract 9 from both sides

-8x = 26

divide both sides by -8

x= -26/8

you can simplify this to -13/4

7 0

3 years ago

Read 2 more answers

Which is greater, 4.87 x 10^9 or 9.212 x 10^8? <br> please helpp

STatiana [176]

Answer: 4.87 x 10^9 is greater than 9.212 *10^8

Step-by-step explanation:

4.87 x 10^9 is greater than 9.212 *10^8 because it has the greater exponent.

4.87 * 10^9 = 4,870,000,000

9.212 *10^8= 921,200,000

4,870,000,000 > 921,200,000

6 0

3 years ago

Can you please help me solve for the area

ycow [4]

Answer:

area is equal to 198 cm^2

Step-by-step explanation:

the height h of the left triangle can be obtained by

Pytagoras theorem:

$5^2+h^2=13^2\\25+h^2=169\\h^2=169-25\\h^2=144\\h=\sqrt{144} \\h=12$

the, the area of the left triangle is

$A_1=\frac{5*12}{2} \\A_1=5*6\\A_1=30$

the area of the right triangle is

$A_2=\frac{2*12}{2} \\A_2=12$

and the area of the rectangle in the middle is

$A_3=(13)(12)\\A_3=156$

Por tanto, el area total sera

$A=A_1+A_2+A_3\\A=30+12+156\\A=198$

8 0

3 years ago

The temperature was -13 and then rose 9. What is the temperature now?

gulaghasi [49]

Answer:

-4

Step-by-step explanation:

bro thats 6th grade math lol and its SO easy just need to try

4 0

3 years ago