A) Variance = 10.24
B) Standard Deviation = 3.2
One of the measurements of dispersion is the standard deviation, which is exclusively used for quantitative data. It aids in determining if the data's mean is a suitable measurement to reflect the core value.
TIME FREQUENCY(f) MIDPOINT(x) d d² fd fd²
0 - 0.9 43 0.45 -3 9 -129 387
1.0 - 1.9 17 1.45 -2 4 -34 68
2.0 - 2.9 19 2.45 -1 1 -19 19
3.0 - 3.9 18 3.45 0 0 0 0
4.0 - 4.9 14 4.45 1 1 14 14
5.0 -5.9 16 5.45 2 4 32 64
∑f = 127
∑fd = -136
∑fd² = 552
Standard Deviation = 
√4.35 - √1.15
Standard Deviation = 3.2
(SD)² = (3.2)² = 10.24
Variance = 10.24
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Answer:
36 runners finished the race.
Step-by-step explanation:
In the first half of the race, 1/4 of them dropped, so we need to calculate 1/4 of 80:
(1/4) * 80 = 80/4 = 20
If 20 runners dropped out, now we have 80 - 20 = 60 runners.
Then, in the second half, 2/5 of the remaining runners dropped out, so:
(2/5) * 60 = 2*60/5 = 120/5 = 24
If 24 runners dropped out, now we have 60 - 24 = 36 runners.
So, 36 runners finished the race.
I don't know what the "six-step method" is supposed to be, so I'll just demonstrate the typical method for this problem.
Let <em>x</em> be the amount (in gal) of the 50% antifreeze solution that is required. The new solution will then have a total volume of (<em>x</em> + 60) gal.
Each gal of the 50% solution used contributes 0.5 gal of antifreeze. Similarly, each gal of the 30% solution contributes 0.3 gal of antifreeze. So the new solution will contain (0.5 <em>x</em> + 0.3 * 60) gal = (0.5 <em>x</em> + 18) gal of antifreeze.
We want the concentration of antifreeze to be 40% in the new solution, so we need to have
(0.5 <em>x</em> + 18) / (<em>x</em> + 60) = 0.4
Solve for <em>x</em> :
0.5 <em>x</em> + 18 = 0.4 (<em>x</em> + 60)
0.5 <em>x</em> + 18 = 0.4 <em>x</em> + 24
0.5 <em>x</em> - 0.4 <em>x</em> = 24 - 18
0.1 <em>x</em> = 6
<em>x</em> = 6/0.1 = 60 gal
