3 years ago

I need help ASAP there’s a time limit

Mathematics

1 answer:

andreyandreev [35.5K]3 years ago

3 0

14 uehebeudygwgwvbsbbe

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Sean asked 10 of his classmates the amount of their weekly allowance. He recorded the information in this table. $15 $15 $22 $14

Anarel [89]

The interquartile range = Third quartile - First quartile = 22 - 15 = 7

The range = 25 - 12 = 13

<h3>What is the Interquartile Range (IQR) and Range ofa Data?</h3>

Range = largest data value - the least data value

Interquartile range = Q3 - Q1

Given the following weekly allowance of Sean's classmates as, 15, 15, 22, 14, 12, 24, 16, 25, 15, 15:

Third quartile (Q3) = 22

First quartile (Q1) = 15

Interquartile range = 22 - 15 = 7

Range = 25 - 12 = 13

Learn more about trhe interquartile range on:

brainly.com/question/4102829

#SPJ1

8 0

2 years ago

1. mutliply 3.and 3/5 enter your answer as an improper fraction in simplified form in the box.

natali 33 [55]

Answer:

1. $\frac{9}{5}$

2. $-11\frac{1}{4}$

3. $10\frac{1}{8}$

4. $4\frac{17}{24}$

5. $-4\frac{7}{12}$

Step-by-step explanation:

In order to solve them you just have to multiply them and add them:

$3 (\frac{3}{5} )=\frac{15+3}{5}=\frac{18}{5}$

$-4\frac{1}{2} (2\frac{1}{2} )=-(\frac{9}{2} )(\frac{5}{2}) =-(\frac{45}{4}) =-11\frac{1}{4}$

$(-2\frac{1}{4} )(-4\frac{1}{2}) =(-\frac{9}{4}) (-\frac{9}{2} )=-\frac{81}{8} =-10\frac{1}{8}$

$3\frac{3}{8}+ 4\frac{1}{3}=7\frac{8+9}{24} =7\frac{17}{24}$

$-2\frac{1}{4} +-2\frac{1}{3}=-4\frac{7}{12}$

5 0

3 years ago

HELP ME OUT PLEASE!!

Zepler [3.9K]

Answer:

may sagot na pala ok bye thanks sa pts

8 0

2 years ago

Read 2 more answers

If you can figure everything out that amazing,but one by one is alright as well

Kay [80]

Answer:

x=14°

SEE the IMAGE FOR SOLUTION

8 0

3 years ago

Y + 4x = -1

Georgia [21]

Hope this helps

3,-13
-1,3
0,-1
3,-13

7 0

3 years ago