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:
The solution of the given system is (x,y) = (1,4)
Step-by-step explanation:
Here, the given system of equation is:
x = -3 + y
26 x - 5 y = 6
Now, substitute x = -3 + y in the second equation , we get
26 x - 5 y = 6 ⇒ 26 (-3 + y) - 5 y = 6
or, 26(-3) + 26(y) - 5y = 6
⇒ - 78 + 26 y - 5y = 6
⇒ 21 y = 84
⇒ y = 84 / 21 = 4 or , y = 4
So x = -3 + y = -3 + 4 = 1
Hence, the solution of the given system is (x,y) = (1,4)
32. Answer: 2+5×3=17
33. Answer: 2×5+3=17
34. Answer: 2×5×3=30
35. Answer: 2×5-3=7