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musickatia [10]
3 years ago
5

The table shows the number of hours that a group of teammates spent in their first week of training for the track finals. In the

second week, they each add 6 hours to their training times. What are the mean,median,mode, and range of times for the second week? 10,8,12,6,5,4,10
Mathematics
1 answer:
Lena [83]3 years ago
6 0

Answer:

Mean:  13.8571

Median:  13

Mode:   16

Range:    8

Step-by-step explanation:

We are given set of entries in that the number of hours taken by a group of teammates spent in their first week of training is:

10            8            12             6             5            4              10

Now in the second week the each of the entry is increased by 6 so the data for the second week is given as:

16             14            18            12              11             10             16

Now, the Mean of this data is given as:

Mean=(sum of all the entries)/(number of entries)

Mean=\dfrac{16+14+18+12+11+10+16}{7}=\dfrac{97}{7}=13.8571

Hence, Mean is 13.8571.

also when we arrange our data in the ascending order with frequency we could easily find our median

 Time spent(in hours)                                  Frequency

     10                                                                 1

      11                                                                  1

      12                                                                 1

      14                                                                  1

       16                                                                 2

       18                                                                 1

Median is the central tendency of a data. As there are two values in the middle i.e. 12 and 14

Hence, the Median will lie between the two i.e. the median will be 13.

Now, the mode of the data is the value corresponding to the highest frequency.

so Mode is 16.

Now, the range of a data is the difference between the highest and the smallest value of the data.

Hence, Range=18-10=8.


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Step-by-step explanation:

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a(t)=\frac{dv}{dt}=\frac{d}{dt} (\frac{105t}{6t+17})

\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'\\\\105\frac{d}{dt}\left(\frac{t}{6t+17}\right)\\\\\mathrm{Apply\:the\:Quotient\:Rule}:\quad \frac{d}{{dx}}\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \frac{{\frac{d}{{dx}}f\left( x \right)g\left( x \right) - f\left( x \right)\frac{d}{{dx}}g\left( x \right)}}{{g^2 \left( x \right)}}

105\cdot \frac{\frac{d}{dt}\left(t\right)\left(6t+17\right)-\frac{d}{dt}\left(6t+17\right)t}{\left(6t+17\right)^2}\\\\105\cdot \frac{1\cdot \left(6t+17\right)-6t}{\left(6t+17\right)^2}\\\\a(t)=\frac{1785}{\left(6t+17\right)^2}

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a(5)=\frac{1785}{\left(6(5)+17\right)^2}=\frac{1785}{2209}\approx0.808 \frac{ft}{s^2}

b. At t = 10 s

a(10)=\frac{1785}{\left(6(10)+17\right)^2}=\frac{255}{847}\approx0.301 \frac{ft}{s^2}

c. At t = 20 s

a(20)=\frac{1785}{\left(6(20)+17\right)^2}=\frac{1785}{18769}\approx0.095 \frac{ft}{s^2}

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