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Leviafan [203]
3 years ago
12

What are the minimum, first quartile, median, third quartile, and maximum of the data set?

Mathematics
1 answer:
allochka39001 [22]3 years ago
4 0
If we rearrange the given data set from lowest to highest, we have 
     3,\:6,\:7,\:8,\:10,\:12,\:15,\:18

The minimum is 3.

The first quartile is the average of 6 and 7, which is 6.5

The median is the average of 8 and 10 which is 9.

The third quartile is the average of 12 and 15 which is 13.5.

The maximum is 18. 
You might be interested in
Which percentage is equivalent to 14/25
Leviafan [203]

46% is equal to 14/25

14/25 is equal to 46/100 (multiply 14/25 times 4/4) and that is 46%

5 0
2 years ago
Read 2 more answers
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DaniilM [7]
Approximately there are some of these on the we
8 0
2 years ago
Z varies jointly with x, and y, and z=7 when x= 2, y= 2
fomenos

Answer:

56

Step-by-step explanation:

Complete question:

Z varies jointly with x and y , x=2 and y=2, z=7. Find z when x=4 and y=8 using joint variation . (I need the problem worked out step by step)

If z varies jointly with x, and y, this is expressed as;

z = kxy

If z=7 when x= 2, y= 2

7  = k(2)(2)

7 = 4k

k = 7/4

To get z when x= 4 and y = 8

z = kxy

z =7/4 (4)(8)

z = 7*8

z = 56

Hence the value of z is 56

7 0
3 years ago
. He rowed 3 1/2 miles in 1/2 hour.What is the average speed in miles per hour.
Rudiy27

Answer:

7 miles per hour

Step-by-step explanation:

3 1/2 miles = 7/2 miles

1/2 hours= 30 minutes

So,

In 30 mins, travelled 7/2 miles

In 1 min, travelled 7/2 ÷ 30= 7/2× 1/30 = (7×1)/(2×30) = 7/60 miles

In 60 mins, travelled (7×60)/60 miles = 420/60 miles = 7 miles.

Hence, the speed is 7 miles per hour.

Hope this helps :)

7 0
2 years ago
3. The backyard of a home is a rectangle 25m by 30m. A garden of uniform width is to be
KiRa [710]

Answer:

Part A

The width of the garden is approximately 3.987 meters

Part B

Please find attached the drawing of the solution

Step-by-step explanation:

Part A

The given parameters are;

The dimensions of the backyard of the home = 25 m by 30 m

The width of the garden to be built in the yard = Uniform width

The shape of the grass left inside = Rectangular

The area of the grass at the center = The area of the garden

Let 'x' represent the width of the garden, we have;

The length of the rectangular grass area at the center, l = 30 - 2·x

The width of the rectangular area of grass at the center, w = 25 - 2·x

The area of the rectangular backyard, A = 30 m × 25 m = 750 m²

The area of the rectangular backyard, A = (The area of the garden) + (The area of the rectangle of grass inside)

The area of the rectangle of grass, GA = (30 - 2·x)·(25 - 2·x) = The area of the garden

The area of the rectangular backyard, A = 750 = (30 - 2·x)·(25 - 2·x) + (30 - 2·x)·(25 - 2·x)  = 2 × (30 - 2·x)·(25 - 2·x) = 8·x² - 220·x + 1,500

∴ 750 = 8·x² - 220·x + 1,500

8·x² - 220·x + 1,500 - 750 = 0

8·x² - 220·x + 750 = 0

x = (220 ± √(220² - 4 × 8 × 750))/(2 × 8)

x ≈ 23.513, or x = 3.987

When x = 25.513, the width of the rectangle of grass inside, w = 25 - 2 × 23.513 = -22.026, which is not a natural (physically possible)

Therefore, the possible width of the garden, x ≈ 3.987

Part B

The drawing of the solution created with MS Visio is attached

6 0
2 years ago
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