178+63+37=278 So if we were ESTIMATING it would be around 280.
I hope that this helps!
From the z-table and at p67 or 0.67 (the decimal values in the table),
Z =0.44
See the attached photo (at the point where 0.6700 appears)
Answer:
0.293 s
Step-by-step explanation:
Using equations of motion,
y = 66.1 cm = 0.661 m
v = final velocity at maximum height = 0 m/s
g = - 9.8 m/s²
t = ?
u = initial takeoff velocity from the ground = ?
First of, we calculate the initial velocity
v² = u² + 2gy
0² = u² - 2(9.8)(0.661)
u² = 12.9556
u = 3.60 m/s
Then we can calculate the two time periods at which the basketball player reaches ths height that corresponds with the top 10.5 cm of his jump.
The top 10.5 cm of his journey starts from (66.1 - 10.5) = 55.6 cm = 0.556 m
y = 0.556 m
u = 3.60 m/s
g = - 9.8 m/s²
t = ?
y = ut + (1/2)gt²
0.556 = 3.6t - 4.9t²
4.9t² - 3.6t + 0.556 = 0
Solving the quadratic equation
t = 0.514 s or 0.221 s
So, the two time periods that the basketball player reaches the height that corresponds to the top 10.5 cm of his jump are
0.221 s, on his way to maximum height and
0.514 s, on his way back down (counting t = 0 s from when the basketball player leaves the ground).
Time spent in the upper 10.5 cm of the jump = 0.514 - 0.221 = 0.293 s.
Increasing 50 by x% can be written as:
.................(1)
Decreasing 70 by x % can be represented by the expression:
...........(2)
Now we are given in the question that if we increase 50 by x% and 70 by x%, then the two values are equal.
so equating (1) and (2) , we get
bringing 0.01x to the left side:
taking 50 to the right side, subtracting it from 70
dividing both sides by 0.02
Answer x=2000
