(a) If a kitten weighs 99 grams at birth, it is at 5.72 percentile of the weight distribution.
(b) For a kitten to be at 90th percentile, the minimum weight is 146.45 g.
<h3>
Weight distribution of the kitten</h3>
In a normal distribution curve;
- 2 standard deviation (2d) below the mean (M), (M - 2d) is at 2%
- 1 standard deviation (d) below the mean (M), (M - d) is at 16 %
- 1 standard deviation (d) above the mean (M), (M + d) is at 84%
- 2 standard deviation (2d) above the mean (M), (M + 2d) is at 98%
M - 2d = 125 g - 2(15g) = 95 g
M - d = 125 g - 15 g = 110 g
95 g is at 2% and 110 g is at 16%
(16% - 2%) = 14%
(110 - 95) = 15 g
14% / 15g = 0.93%/g
From 95 g to 99 g:
99 g - 95 g = 4 g
4g x 0.93%/g = 3.72%
99 g will be at:
(2% + 3.72%) = 5.72%
Thus, if a kitten weighs 99 grams at birth, it is at 5.72 percentile of the weight distribution.
<h3>Weight of the kitten in the 90th percentile</h3>
M + d = 125 + 15 = 140 g (at 84%)
M + 2d = 125 + 2(15) = 155 g ( at 98%)
155 g - 140 g = 15 g
14% / 15g = 0.93%/g
84% + x(0.93%/g) = 90%
84 + 0.93x = 90
0.93x = 6
x = 6.45 g
weight of a kitten in 90th percentile = 140 g + 6.45 g = 146.45 g
Thus, for a kitten to be at 90th percentile, the approximate weight is 146.45 g
Learn more about standard deviation here: brainly.com/question/475676
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Answer:
F=1.47 KN
Explanation:
Given that
Diameter of plate = 25 cm
Height of pool h = 3 m
We know that force can be given as
F= P x A
P=ρ x g x h
Now by putting the values
P=1000 x 10 x 3
P= 30 KPa


F= 30 x 0.049 KN
F=1.47 KN
So the force on the plate will be 1.47 KN.
Answer:
a₁= 1.98 m/s² : magnitud of the normal acceleration
a₂=0.75 m/s² : magnitud of the tangential acceleration
Explanation:
Formulas for uniformly accelerated circular motion
a₁=ω²*r : normal acceleration Formula (1)
a₂=α*r: normal acceleration Formula (2)
ωf²=ω₀²+2*α*θ Formula (3)
ω : angular velocity
α : angular acceleration
r : radius
ωf= final angular velocity
ω₀ : initial angular velocity
θ : angular position theta
r : radius
Data
r =0.4 m
ω₀= 1 rad/s
α=0.3 *θ , θ= 2π
α=0.3 *2π= 0,6π rad/s²
Magnitudes of the normal and tangential components of acceleration of a point P on the rim of the disk when theta has completed one revolution.
We calculate ωf with formula 3:
ωf²= 1² + 2*0.6π*2π =1+2.4π ²= 24.687
ωf=
=4.97 rad/s
a₁=ω²*r = 4.97²*0.4 = 1.98 m/s²
a₂=α*r = 0,6π * 0.4 = 0.75 m/s²