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pashok25 [27]
2 years ago
14

I NEED HELP ASAP PLEASE TEST QUESTIONS solve for X​

Mathematics
1 answer:
Elenna [48]2 years ago
3 0

Answer: x=6

Step-by-step explanation:

x-3+2x-5=10

3x-8       =10

      3x    = 10+8

        x     = 18/3

        x      = 6

Hope this help! don't forget to click "Thanks"! :)

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Consider the sequence 3, -9, 27, -81, .. Find the 14th term of the sequence.
bagirrra123 [75]

Answer:

let 'a' be the first term, 'd' be the common difference between all the terms of the sequence

Step-by-step explanation:

therefore, a = 3,

and, d = -9 -3

          = -12

hence the 14th term would be,

=> a + 13d

=> 3 + 13( -12 )

=> 3 - 156

=> - 153

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A pyramid and a cone are both 10 centimeters tall and have the same
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the area of a rectangular dog pen is 8 1/2 square feet. if the width is 3 2/5 feet, what is the length, in feet?
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8.5 = 3.4 \times x
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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

The first quartile would be 5 since we analyze 1.2, 4.3, 5.0, 5.8, 6.2 and the middle point is 5.0

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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