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Alexandra [31]
3 years ago
8

Ling lives 2 miles from school. It took him 15 minutes to bike from school to home. The first half of the distance he biked at a

speed of 12 mph. What was his speed for the remaining distance? What was his average speed?
Mathematics
2 answers:
kodGreya [7K]3 years ago
8 0
He was going 6 mph for the remaining distance 
the average is 9
lisov135 [29]3 years ago
5 0
1 mile at 12 m/h is .2 m/min
1 mile at .2 m/min (1/.2) = 5 min So the first mile took 5 minutes
To find how fast the second mile was:
You know it was 1 mile and it took 10 minutes
1/10=.1 m/min which is 6 m/h
The average speed is 6+12/2=9 m/h

Hope that helps.  Feel free to ask any questions.
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Which linear equation is represented by the table?<br><br> x −2 1 3 6<br> y 7 4 2 −1
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Step-by-step explanation:

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3 years ago
Identify the lower class​ limits, upper class​ limits, class​ width, class​ midpoints, and class boundaries for the given freque
maw [93]

Answer:

The number of individuals included in the summary is 146.

Step-by-step explanation:

The frequency distribution table provided is as follows:

Class Intervals    Frequency

    100 - 199               24

   200 - 299              90

   300 - 399              27

   400 - 499                1

   500 - 599               4

The lower class limit it the smallest value of each class interval.

Lower class limit = {100, 200, 300, 400, 500}

The upper class limit it the highest value of each class interval.

Upper class limit = {199, 299, 399, 499, 599}

The lower class boundaries are the lower class limits decreased by 0.5 and the upper class boundaries are the upper class limits increased by 0.5.

Class boundaries:

   99.5 - 199.5  

  199.5 - 299.5

  299.5 - 399.5

  399.5 - 499.5

  499.5 - 599.5

The class width is the difference between the class boundaries of each class.

Class width = 199.5 - 99.5 = 100

So, the class width is 100.

The midpoints of a class is the average value of the boundaries of a class.

\text{Midpoint}_{100-199}=\frac{\text{Lower class boundary} + \text{Upper class boundary}}{2}\\

                         =\frac{99.5+199.5}{2}\\\\=149.5

\text{Midpoint}_{200-299}=\frac{\text{Lower class boundary} + \text{Upper class boundary}}{2}\\

                         =\frac{199.5+299.5}{2}\\\\=249.5

\text{Midpoint}_{300-399}=\frac{\text{Lower class boundary} + \text{Upper class boundary}}{2}\\

                         =\frac{299.5+399.5}{2}\\\\=349.5

\text{Midpoint}_{400-499}=\frac{\text{Lower class boundary} + \text{Upper class boundary}}{2}\\

                         =\frac{399.5+499.5}{2}\\\\=449.5

\text{Midpoint}_{500-599}=\frac{\text{Lower class boundary} + \text{Upper class boundary}}{2}\\

                         =\frac{499.5+599.5}{2}\\\\=549.5

The number of individuals included in the summary is the sum of all frequencies.

\text{Number of Individuals}=24 + 90 + 27 + 1 + 4=146

Thus, the number of individuals included in the summary is 146.

3 0
3 years ago
Read 2 more answers
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