Answer:
So then we will have at least 75% of the data within two deviations from the mean .
So then the interval would be (97.14, 99.5)
Explanation:
Previous concepts
Chebyshev’s rule is appropriate for any distribution. "Chebyshev’s inequality applies to all distributions, regardless of shape". And is useful since provides a "minimum percentage of the observations that lies within k standard deviations of the mean. "
If k = 2, at least 3/4 of the measurements lie within 2 standard deviations to within the mean.
And the general formula is (1-1/k^2) represent the fraction of the data within the mean .
Solution to the problem
For this case we want to find the percentage of data that would be at least within two deviations from the mean so for this case the value of k =2 and if we replace we got:
So then we will have at least 75% of the data within two deviations from the mean .
For the other part we have the mean and deviation provided the interval would be:
So then the interval would be (97.14, 99.5)
Answer:
speed is -20.0155 m/s
and direction is downward
Explanation:
given data
mass of rocket = 1500 kg
rocket accelerates upward = 10 m/s²
time t = 2 s
maximum height = 530 m
to find out
speed and direction of the heavier fragment just after the explosion
solution
we use here principle of conservation of momentum that is
Mt Vt=M1 V1 + M2 V2 ................1
so here
first we find the initial velocity that is express as
v = a × t .....................2
v = 10 × 2
v = 20 m/s
and distance S
S =u × t + 0.5 × a × t² ...................3
S = 0 + 0.5 × 10 × 2²
S = 20 m
and now we find speed of the lighter piece that is
v² = v1² + 2 × a × S .................4
0 = v1² + 2 × -9.81 × (530-20)
v1² = 10006.2
v1 = 100.031 m/s
so now put all these in equation 1 we get
Mt Vt = M1 V1 + M2 V2
1500 × 20 = (1500 × 1/3 ) × 100.031 + (1500 × 2/3) + V2
30000 = 50015.5 + 1000 V2
V2 = -20.0155 m/s
so speed is -20.0155 m/s
that mean vertical component of second fragment will be move downward
Answer:
3.185 x 10^-3 cm/s
Explanation:
dV / dt = 4 cubic cm per second
r = 10 cm
The volume of sphere is given by
V = 4/3 x π x r³
Differentiate both sides with respect to t
dV / dt = 4/3 x π x 3r² x dr/dt
Put dV / dt = 4 cubic cm per second, r = 10 cm
4 = 4/3 x 3.14 x 3 x 10 x 10 x dr/dt
dr/dt = 3.185 x 10^-3 cm/s
Answer:
(a) The least time required for the rotation is 50 seconds
(b) The corresponding value of α₁ is 1.6 rad/s²
Explanation: Please see the attachments below
A recreational flyer does not need an airspace authorization in uncontrolled airspace.
An airspace in which the Air Traffic Control service is not considered necessary is called uncontrolled airspace.
In general, an uncontrolled airspace remains at 400 or under 400 levels surpassing the ground.
Moreover, in an uncontrolled airspace the recreational flyer does not need any prior authorization and should also have particular knowledge about the airport traffic patterns
To put it simply, the remote pilots and flyers are not allowed to use landing and take off areas in an uncontrolled airspace.
If you need to learn more about airspace click here:
brainly.com/question/14451166
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