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zepelin [54]
3 years ago
13

4" alt=" - \frac{1}{2} x > 4" align="absmiddle" class="latex-formula">

​
Mathematics
1 answer:
elena-s [515]3 years ago
5 0

Answer:

x< -8

Step-by-step explanation:

To solve the inequality, we must get x isolated.

-1/2x>4

We have to perform the inverse operation to both sides. x is being multiplied by -1/2. The inverse of multiplication is division. Divide both sides by -1/2.

-1/2x / -1/2 > 4/ -1/2

When dividing by fraction, you can also multiply by the reciprocal.

To find the reciprocal, flip the numerator and the denominator.

-1/2–> -2/1

-2/1 *-1/2x > 4* -2/1

x> 4* -2/1

When multiplying or dividing by a negative number in an inequality, you must flip the inequality sign.

x < 4* -2/1

x < 4*-2

x< -8

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A violin student records the number of hours she spends practicing during each of nine consecutive weeks: 6.2 5.0 4.3 7.4 5.8 7.
spin [16.1K]

Answer:

A) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the first  quartile.

Correct we satisfy that 1.2 <1.7 the lower limit

Step-by-step explanation:

Assuming this complete question: A violin student records the number of hours she spends practicing during each of nine consecutive weeks:

6.2 5.0 4.3 7.4 5.8 7.2 8.4 1.2 6.3

For this case we need to sort the data first on increasing way and we got:

1.2, 4.3, 5.0, 5.8, 6.2, 6.3, 7.2, 7.4, 8.4

For this case we have 9 values the median would be on the 5 position:

Median = 6.2

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The third quartile would be 7.2 since we analyze 6.2, 6.3, 7.2, 7.4, 8.4 and the middle point is 7.2

Then the interquartile rnage would be:

IQR = Q_3 -Q_1= 7.2-5=2.2

And 1.5 IQR = 1.5*2.2=3.3

The lower limit on this case would be:

LL= Q_1 -1.5 IQR= 5-3.3=1.7

And our value is 1.2 is lower than the lower limit 1.7

Considering the smallest data value (1.2) and using the 1.5 × IQR rule, we would:

A) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the first  quartile.

Correct we satisfy that 1.2 <1.7 the lower limit

B) not classify the value 1.2 as an outlier, because it is not more than 1.5 × IQR below  the first quartile.

False 1.2 is more than 1.5 IQR below the first quartile

C) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the  median.

False, we analyze if is an outlier comparing with the Q1 or Q3 not with the median

D) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the  mean.

False we analyze if is an outlier comparing with the Q1 or Q3 not with the mean

5 0
3 years ago
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Step-by-step explanation:

if you look at it carefully the endpoint is at the bottom of the Y value which looks like negative 2 this is the only answer you can be

3 0
3 years ago
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3 years ago
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If a || b  and e|| f
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3 0
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Answer:

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Step-by-step explanation:

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