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Julli [10]
3 years ago
5

5, 18, 6, 18, 13 WHATS THE INTERQUARTILE RANGE??

Mathematics
1 answer:
Harlamova29_29 [7]3 years ago
5 0

Answer: 12.5

Step-by-step explanation:

25th Percentile: 5.5

50th Percentile: 13

75th Percentile: 18

Interquartile Range: 12.5

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What is the value of x?
azamat

Answer:

17 <u>1</u>

3

Step-by-step explanation:

= <u>67</u><u>5</u>+ <u>3</u> x = <u>53</u>

100 8 4

------------------------- × 1000

6,750+ 375x = 13,250

375x = 6,500

x = <u>6</u><u>,</u><u>5</u><u>0</u><u>0</u> = <u>2</u><u>6</u><u>0</u> = <u>5</u><u>2</u> = 17 <u>1</u>

375 15 3 3

4 0
3 years ago
B) The mean of 7, 5, 0, 1, 3, 4 and h is 10. Find the value of h.
wariber [46]

Step-by-step explanation:

mean =10

sum of no =7+5+0+1+3+4+h=20+h

N=7

now

mean=sum of no/N

10=20+h/7

10×7=20+h

70-20=h

h=50

7 0
2 years ago
Could someone please answer this question?
sdas [7]

Answer:

The new area will be 1 1/2 of the old area

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
How many L of A 90% saline solution must be mixed with 4 L of a 50% saline solution to make a 74% solution ?
insens350 [35]

Answer:

6 L

Step-by-step explanation:

Given two solutions:

1st solution is 90% saline solution. Let the volume of this solution be x litres.

2nd solution is 50% saline solution. Volume of this solution = 4 L

Resultant solution is 74% saline.

To find:

The volume of 1st solution = ?

Solution:

Total volume of the 74% saline mixture = (4+x) Litres

We can write equation here, as per the percentage of saline in the mixtures.

90% of x L + 50% of 4 L = 74% of (x+4)

\Rightarrow \dfrac{90x}{100} + \dfrac{50}{100}\times 4 = \dfrac{74}{100}\times (x+4)\\\Rightarrow 90x + 200 = 74x+296\\\Rightarrow 90x-74x = 296-200\\\Rightarrow 16x = 96\\\Rightarrow x = \bold{6\ L}

Therefore, the volume must be <em>6 L</em>.

8 0
3 years ago
Mr.ward and mr. heskey are racing toy cars and Mr. heskey wants to give mr. ward an 8 ft head start because his car is 2 seconds
DiKsa [7]

Answer:

  F.  6s > 4s +8

Step-by-step explanation:

From a reference point at the beginning of the track, Ward's car will be located 4s+8 feet down the track after s seconds. That is, it starts 8 feet down the track, and increases its distance by 4 feet every second.

From the same reference point, Heskey's car will be located 6s feet down the track after s seconds. Its distance starts from zero and is increasing at the rate of 6 feet every second.

For Heskey's distance to exceed Ward's distance you want ...

  6s > 4s +8 . . . . . . matches choice F

6 0
3 years ago
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