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AnnyKZ [126]
3 years ago
14

Which statement about the data is true?

Mathematics
2 answers:
Aleksandr [31]3 years ago
8 0
Your correct b is the correct answer

dalvyx [7]3 years ago
4 0

Answer:

(B)

Step-by-step explanation:

The given data set is:

228, 251, 22, 206, 254, 288, 228, 242

Since, in the given data set, 22 is an outlier which will affect the mean of the data, thus mean cannot be used to measure the center of the data due to the presence of an outlier.

Also, the median is the central value which is used to measure the center of the given data set, thus median can be used to find the measures of center.

The mode can be calculated for qualitative data as well as for quantitative data and can also be used to find the measures of center.

Thus, option B is correct.

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t=-28

Step-by-step explanation:

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Unit Exam: Rational Numbers
Minchanka [31]
Assuming that you’re asking for a fraction of 0.31 where 31 repeats?

31/99
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2 years ago
Which ordered pair makes both inequalities true? y &lt; 3x – 1 y &gt; –x + 4 On a coordinate plane, 2 straight lines are shown.
Katen [24]

Answer:

(4,0)

Step-by-step explanation:

we have

y< 3x-1 ----> inequality A

y \geq -x+4 ----> inequality B

we know that

If a ordered pair is a solution of the system of inequalities, then the ordered pair must satisfy both inequalities (makes true both inequalities)

Verify each ordered pair

case 1) (4,0)

<em>Inequality A</em>

0< 3(4)-1

0< 11 ----> is true

<em>Inequality B</em>

0 \geq -(4)+4

0 \geq 0 ----> is true

so

the ordered pair makes both inequalities true

case 2) (1,2)

<em>Inequality A</em>

2< 3(1)-1

2< 2 ----> is not true

so

the ordered pair not makes both inequalities true

case 3) (0,4)

<em>Inequality A</em>

4< 3(0)-1

4< -1 ----> is not true

so

the ordered pair not makes both inequalities true

case 4) (2,1)

<em>Inequality A</em>

1< 3(2)-1

1< 5 ----> is true

<em>Inequality B</em>

1 \geq -(2)+4

1 \geq 2 ----> is not true

so

the ordered pair not makes both inequalities true

5 0
3 years ago
Read 2 more answers
How to find upper and lower bound of a definite integral given an unknown equation?
worty [1.4K]
Finding the upper and lower bounds for a definite integral without an equation is pretty hard because how can we find the upper and lower bounds of  definite integral if there is no equation given. But I will teach you how to find the lower and upper bounds of a definite integral, when the equation is like this 

\int\limits^6_1 t^{2} - 6t + 11dt 

So, i integrate this, 
( \frac{t^{3} }{3} - 3t^{2} + 11t) \int\limits^6_1

I know I have a minimum at x=3 because;
f(t )= t^2 − 6t + 11
f′(t) = 2
t−6 = 0
2(t−3) = 0
t = 3
f(5) = 4
f(1) = −4
4 0
3 years ago
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