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monitta
2 years ago
14

What is the range for this set of data? 7, 15, 12Mrs. Juarez graded ten English papers and recorded the scores. 92, 95, 100, 62,

88, 90, 100, 96, 89, 98 Which statements are true? Check all that apply.
Mathematics
1 answer:
ra1l [238]2 years ago
5 0

Answer:

Following are the solution to the given question:

Step-by-step explanation:

Please find the complete question in the attached file.

Arrange the value into the ascending order:

62, 88, 89, 90, 92, 95, 96, 98, 100, 100

Calculating the Range:

= Greatest - Lowest\\\\= 100 - 62\\\\ = 38

Outlier =  This might be the value "far" from other data values= 62

The range of scores would be, sans outlier,

 88, 89, 90, 92, 95, 96, 98, 100, 100

Calculating the range without outlier = 100 - 88 = 12

Inter Quartile Range = Q_3 - Q_1

IQR = 98 - 89 = 11

Calculating the IQR without outlier = 99 - 89.5 = 9.5

Therefore,

Option a is "True".

Option b is "True".

Option c is "True".

Option d is "False".

Option e is "False".

Option f is "True".

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12.5 m in 1 second.

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Henry is saving money to buy a game. so far he has saved $12 , which is three-fourths of the total cost of the game. how much do
vazorg [7]
Henry has 3 / 4 0f total cost so,
Let ( a ) is total cost of game

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5 0
3 years ago
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He counted 206 wheels and 170 pedals. How many bicycles and tricycles does he have.
noname [10]
Bicycle has 2 wheels and 2 pedals
 tricycle has 3 wheels and 2 pedals

EQ 1 : 2b + 2t = 170 pedals

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subtract  EQ 1 from EQ 2

2b +3t = 206 - 2b + 2t = 170 = t=36

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170 pedals - 72 = 98 pedals left
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8 0
3 years ago
For what values of $x$ is $$\frac{x^2 + x + 3}{2x^2 + x - 6} \ge 0?$$ note: be thorough and explain why all points in your answe
Rus_ich [418]

To solve the inequality \frac{x^2+x+3}{2x^2+x-6}\ge \:0

\mathrm{Factor\:the\:left\:hand\:side\:}\frac{x^2+x+3}{2x^2+x-6}:

The numerator x^2+x+3 is not factorizable.

so factor the denominator 2x^2+x-6:

2x^2+x-6=\left(2x^2-3x\right)+\left(4x-6\right)=x\left(2x-3\right)+2\left(2x-3\right)\\ \mathrm{Factor\:out\:}x\mathrm{\:from\:}2x^2-3x\mathrm{:\quad }x\left(2x-3\right)\\ \mathrm{Factor\:out\:}2\mathrm{\:from\:}4x-6\mathrm{:\quad }2\left(2x-3\right)\\

Now take \mathrm{Factor\:out\:common\:term\:}\left(2x-3\right)

Then we get factor of the denominator as \left(2x-3\right)\left(x+2\right)

Thus \frac{x^2+x+3}{2x^2+x-6}=\frac{x^2+x+3}{\left(x+2\right)\left(2x-3\right)}

Now \mathrm{Compute\:the\:signs\:of\:the\:factors\:of\:}\frac{x^2+x+3}{\left(x+2\right)\left(2x-3\right)}\\

Signs of x^2+x+3>0

\mathrm{Choosing\:ranges\:that\:satisfy\:the\:required\:condition:}\:\ge \:\:0

x\frac{3}{2} is the required solution of the given inequality.

5 0
3 years ago
Read 2 more answers
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