The following stem-and-leaf plot represents the times in minutes required for 26 co-workers to commute to work. Use the data provided to find the quartiles. Commute Times in Minutes Stem Leaves 1 0 4 4 6 6 6 9 2 2 4 4 5 8 3 3 4 4 5 6 8 4 0 0 3 3 5 6 7 8 Key: 1|0

Q1 = 18.25 minutes

Q2 = 33.5 minutes

Q3 = 40.75 minutes

Step-by-step explanation:

The stem and leaf diagram is:

Stem | Leaves

1         |  0 4 4 6 6 6 9

2        |  2 4 4 5 8

3        |  3 4 4 5 6 8

4        |  0 0 3 3 5 6 7 8

We need to find the quartiles Q1, Q2 and Q3. We know that,

Q1 = (1/4)*(n+1)th value

Q2 = (1/2)*(n+1)th value

Q3 = (3/4)*(n+1)th value

where n is the total number of co-workers. So,

Q1 = (1/4)*(26+1)th value

Q1 = 6.75th value

we need to count the leaves in the plot starting from the first one until we reach the 6.75th value. So, by counting we conclude that the 6.75th value lies between the 6th and 7th value i.e. 16 and 19.

Q1 = 16 + (19-16)*0.75

     = 16 + 2.25

Q1 = 18.25 minutes

Q2 = (1/2) * (26+1)th value

     = 13.5th value.

From the plot we find that the 13.5th value lies in the middle of the 13th and 14th value i.e. 33 and 34. So,

Q2 = (33+34)/2

     = 67/2

Q2 = 33.5 minutes

Q3 = (3/4)*(26+1)th value

     = 20.25th value

From the plot we find that the 20.25th value lies somewhere between the 20th and 21st value i.e. 40 and 43. So,

Q3 = 40 + (43-40)*0.25

     = 40 + 0.75

Q3 = 40.75 minutes

c

step-by-step explanation:

it isn't d, because the interquartile range is from 27-57. and the interquartile range is basically half the data, it's pretty obvious.

it isn't b, because like i said earlier, the interquartile range is from 27-57. 57-27 is 30. the real range has to be longer

not a either. although it looks like it, 3/4 of the data is not greater than 27. but if c is wrong than this should be correct

the correct answer is c, because 13 is the point in which the plot starts

infinitely many solutions

step-by-step explanation:

we have been given the equation;

2(2x + 3) = -4 + 2(2x + 5)

we need to determine the value of x.

2(2x + 3), we expand this expression to obtain 4x + 6

similarly 2(2x + 5) will become 4x + 10

now we have;

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 10 - 4

4x + 6 = 4x + 6

the expression on the left hand side is identical to the one on the right hand side. therefore, we are looking at just a single straight line whose equation is;

y = 4x + 6

consequently, we shall have infinitely many solutions