Histograms are useful when we have data which can be divided into several classes or groups. The histogram shows the trend of each class and the trend among the different classes. For example when we have about 50 different values ranging from 1 to 20, it will be a better approach to draw a histogram in this case by dividing the data into small ranges e.g 1 to 4, 5 to 9 and so on and counting the frequency for each class.
Dot plot is useful when we have a small number of individual values. In this case we can visualize how many times each individual value occurred in the data. This is useful when the number of values in the data is less.
In the given scenario, we have 12 values in total ranging from 1 to 5. So making a dot plot would be the best choice. A histogram would not be useful in this case.
Therefore, the correct answer is option D. Dot plot, because a small number of scores are reported individually
tan2x*cotx - 3 = 0
We know that: tan2x = sin2x/cos2x and cotx = cosx/sinx
==> sin2x/cos2x *cosx/sinx = 3
Now we know that sin2x = 2sinx*cosx
==> 2sinxcosx/cos2x * cosx/sinx = 3
Reduce sinx:
==> 2cos^2 x/ cos2x = 3
Now we know that cos2x = 2cos^2 x-1
==> 2cos^2 x/(2cos^2 x -1) = 3
==> 2cos^2 x = 3(2cos^2 x -1)
==> 2cos^2 x = 6cos^2 x - 3
==> -4cos^2 x= -3
==> 4cos^2 x = 3
==> cos^2 x = 3/4
==> cosx = +-sqrt3/ 2
<span>==> x = pi/6, 5pi/6, 7pi/6, and 11pi/6</span>
Answer:a) P(8 of the players numbers are drawn)=1.3×10^-8
b) P(7 of the players number are drrawn)=3.33×10^-c) P(at least 6 of the players number were drawn)=1.84×10^-4
Step-by-step explanation:
Players has 8 combinations of numbers from 1-40. The outcome S contains all the combinations of 8 out of 40
a) P(8 of the players numbers are drawn)= 1/40/8= 1.3×10^-8
There are one in hundred million chances that the draw numbers are precisely the chosen ones.
b) Number of ways of drawing 78 selected numbers from 1-40=8×(40-7)
8×32
P(7 of the players number are drawn)=8×32/40 =3.33×10^-6.
There are approximately 300,000 chances that 7 of the players numbers are chosen
c) P(at least 6 players numbers are drawn)= 32/2×(8/6) ways to draw.
P(at least 6 players numbers are drawn)=P(all 8 chosen are drawn)+P(7 players numbers drawn)+P(6 chosen are drawn) = 1+ 8 x32/40/8 +[8\6 ×32/2]
P(at least 6 players numbers are drawn) = 1.84×10^-4.
There are approximately 5400chances that at least6 of the numbers drawn are chosen by the player.