the data represents the heights of fourteen basketball players, in inches. 69, 70, 72, 72, 74, 74, 74, 75, 76, 76, 76, 77, 77, 8
mamaluj [8]
If you would like to know the interquartile range of the new set and the interquartile range of the original set, you can do this using the following steps:
<span>The interquartile range is the difference between the third and the first quartiles.
The original set: </span>69, 70, 72, 72, 74, 74, 74, 75, 76, 76, 76, 77, 77, 82
Lower quartile: 72
Upper quartile: 76.25
Interquartile range: upper quartile - lower quartile = 76.25 - 72 = <span>4.25
</span>
The new set: <span>70, 72, 72, 74, 74, 74, 75, 76, 76, 76, 77, 77
</span>Lower quartile: 72.5
Upper quartile: 76
Interquartile range: upper quartile - lower quartile = 76 - 72.5 = 3.5
The correct result would be: T<span>he interquartile range of the new set would be 3.5. The interquartile range of the original set would be more than the new set.</span>
For simple interest, just multiply all of them together
note that 2% = 0.02
570(0.02)(4) = 45.6
simple interest is 45.6
oh and if you need the equation, it is p(r)t, or principle x rate x time
hope this helps
The quotient is 10.6 the answer is B.
Answer:
sin^2 A
Step-by-step explanation:
first expand the trigonometry
1(1-cosA)+cosA(1-cosA)
=1-cosA+cosA-cos^2 A
= 1-cos^2 A
from trigonometric identity sin^2 A + cos^2 A= 1
sin^2 A= 1- cos^2 A
=sin^2 A
<span>Acceleration of a passenger is centripetal acceleration, since the Ferris wheel is assumed at uniform speed:
a = omega^2*r
omega and r in terms of given data:
omega = 2*Pi/T
r = d/2
Thus:
a = 2*Pi^2*d/T^2
What forces cause this acceleration for the passenger, at either top or bottom?
At top (acceleration is downward):
Weight (m*g): downward
Normal force (Ntop): upward
Thus Newton's 2nd law reads:
m*g - Ntop = m*a
At top (acceleration is upward):
Weight (m*g): downward
Normal force (Nbottom): upward
Thus Newton's 2nd law reads:
Nbottom - m*g = m*a
Solve for normal forces in both cases. Normal force is apparent weight, the weight that the passenger thinks is her weight when measuring by any method in the gondola reference frame:
Ntop = m*(g - a)
Nbottom = m*(g + a)
Substitute a:
Ntop = m*(g - 2*Pi^2*d/T^2)
Nbottom = m*(g + 2*Pi^2*d/T^2)
We are interested in the ratio of weight (gondola reference frame weight to weight when on the ground):
Ntop/(m*g) = m*(g - 2*Pi^2*d/T^2)/(m*g)
Nbottom/(m*g) = m*(g + 2*Pi^2*d/T^2)/(m*g)
Simplify:
Ntop/(m*g) = 1 - 2*Pi^2*d/(g*T^2)
Nbottom/(m*g) = 1 + 2*Pi^2*d/(g*T^2)
Data:
d:=22 m; T:=12.5 sec; g:=9.8 N/kg;
Results:
Ntop/(m*g) = 71.64%...she feels "light"
Nbottom/(m*g) = 128.4%...she feels "heavy"</span>
