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Tomtit [17]
3 years ago
13

What is the common difference of an arithmetic sequence defined by the general formula: an=4n+5

Mathematics
1 answer:
FromTheMoon [43]3 years ago
4 0

Answer:

4

Step-by-step explanation:

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4. This box plot shows the number of marks earned by the boys in a class test. What was the lower quartile? 5.This box plot show
suter [353]

Answer:

4-lower- 18

5-upper-29

6- range-56

Step-by-step explanation:

If you look up a completed box plot you can tell where lower and upper quartiles are. Range is found by Max-Min

Hope I helped...!

4 0
3 years ago
"When I started here, there were 500 employees. Since then, we have
Len [333]

Answer:

Assuming that the question is asking how many employees there are, your answer would be 675.

Step-by-step explanation:

  1. First, to find 35% of 500, you can use this equation (35 x 500) / 100
  2. Second, solve the equation (The answer is 175)
  3. Third, you add 500 + 175 to get 675
4 0
2 years ago
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
Make z the subject 9d= 6z - 5d
yarga [219]

Answer:

z=2\frac{1}{3}d

Step-by-step explanation:

9d=6z-5d

6z=9d+5d

6z=14d (combine like terms)

z=2\frac{1}{3}d

Hope this helps :)

4 0
3 years ago
What is the simplified form of each expression?<br>1.7m^2 + 6.5n - 4n + 2.5m^2 - n
Ulleksa [173]

{4.2m}^{2}  + 1.5n
3 0
3 years ago
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