We have x = (2/5)y and x + y = 35;
then, (2/5)y + y = 35;
2y + 5y = 175;
7y = 175;
y = 27;
x = 35 - 27;
x = 8;
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:
Changing the net force on an object can either impress on it even more force in the direction that it is moving, making it go faster, or if you apply more net force on the object against it, then it decreases acceleration
Step-by-step explanation:
I suppose that when you write "log^6" , you want to say log₆
Change of base formula:
loga b=logc b / logc a
log₆ 120=ln 120 / ln 6=2.67
Answer: log₆ 120=2.67