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Tamiku [17]
3 years ago
9

Alexa needs to order Seventeen-fifths, 3Three-fourths, and StartFraction 19 over 20 EndFraction from least to greatest. Which st

atements about those fractions are true? Check all that apply. StartFraction 19 over 20 EndFraction is less than 3Three-fourths. 3Three-fourths is greater than Seventeen-fifths. Seventeen-fifths is the largest fraction shown. Two of the fractions shown are equivalent fractions. 20 is the common denominator of all 3 fractions. . Helpppppp
Mathematics
1 answer:
masya89 [10]3 years ago
6 0

Answer:

Step-by-step explanation:

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54; 18; 76; 42; 76; and 46<br> mean: <br><br><br> median: <br><br><br> mode:
Leokris [45]
To work out the mean, add them all up and divide by however many values there are:

(18+54+42+(76x2)+46) / 6 = 52

To work out the median, order them  from largest to smallest or vice versa and find the middlest value.

18, 42, 46, 54, 76 ,76

(46 + 54) / 2 = 50

The mode is the most frequently occurring number, so it would be 76. 

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4 0
3 years ago
Question content area top
klasskru [66]

Answer:

12

Step-by-step explanation:

if we know we want a profit of at least 114, we need to solve for x.

if we put 12 for x we get the answer 117 and when we put 11 for x we get less than 114.

-0.5(12)^2 + 25(12)-111 = 117

3 0
2 years ago
A class sold tickets to their school play. Each of the 23 students in the class sold 4 tickets. The cost of each
11111nata11111 [884]

Answer:

$644

Step-by-step explanation:

7 × 4 = 28

28 × 23 = 644

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8 0
3 years ago
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Plz help me with this question!!!
Alona [7]

Answer:

6.111111111111111111111111111111111111111111111111111 or 6.11 repeating

Step-by-step explanation:

7 0
3 years ago
If JK=20-x^2, KL=2-x, and JL=10 find x
kherson [118]
|JK|=20-x^2;\ |KL|=2-x;\ |JL|=10\\D:20-x^2 > 0\ and\ 2-x > 0\ and\ 20-x^2 < 10\ and\ 2-x < 10\\x\in(-\sqrt{20};\ \sqrt{20})\ and\  x < 2\ and\\x\in(-\infty; -\sqrt{10})\ \cup\ (\sqrt{10};\ \infty)\ and\ x >-8\\\\x\in(-\sqrt{20};-\sqrt{10})\\\\|JL|=|JK|+|KL|\\\\20-x^2+2-x=10\\\\-x^2-x+22=10\ \ \ \ \ |subrtact\ 10\ from\ both\ sides\\\\-x^2-x+12=0\\\\-(x^2+x-12)=0\\\\x^2+x-12=0\\\\x^2+4x-3x-12=0\\\\x(x+4)-3(x+4)=0\\\\(x+4)(x-3)=0\iff x+4=0\ or\ x-3=0\\\\x=-4\in D\ or\ x=3\notin D\\\\Answer:\boxed{x=-4}
6 0
3 years ago
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