Answer:
Answer:
Q_1 = 7Q
1
=7
Q_2 = 10Q
2
=10
Q_3 = 13.5Q
3
=13.5
Step-by-step explanation:
Given
5, 7, 7, 8, 10, 11, 12, 15, 17.
Required
Determine Q1, Q2 and Q3
The number of data is 9
Calculating Q1:
Q1 is calculated as:
Q_1 = \frac{1}{4}(N + 1)Q
1
=
4
1
(N+1)
Substitute 9 for N
Q_1 = \frac{1}{4}(9 + 1)Q
1
=
4
1
(9+1)
Q_1 = \frac{1}{4}*10Q
1
=
4
1
∗10
Q_1 = 2.5th\ itemQ
1
=2.5th item
This means that the Q1 is the mean of the 2nd and 3rd data.
So:
Q_1 = \frac{1}{2}(7+7)Q
1
=
2
1
(7+7)
Q_1 = \frac{1}{2}*14Q
1
=
2
1
∗14
Q_1 = 7Q
1
=7
Calculating Q2:
Q2 is calculated as:
Q_2 = \frac{1}{2}(N + 1)Q
2
=
2
1
(N+1)
Substitute 9 for N
Q_2 = \frac{1}{2}(9 + 1)Q
2
=
2
1
(9+1)
Q_2 = \frac{1}{2}*10Q
2
=
2
1
∗10
Q_2 = 5th\ itemQ
2
=5th item
Q_2 = 10Q
2
=10
Calculating Q3:
Q3 is calculated as:
Q_3 = \frac{3}{4}(N + 1)Q
3
=
4
3
(N+1)
Substitute 9 for N
Q_3 = \frac{3}{4}(9 + 1)Q
3
=
4
3
(9+1)
Q_3 = \frac{3}{4}*10Q
3
=
4
3
∗10
Q_3 = 7.5th\ itemQ
3
=7.5th item
This means that the Q3 is the mean of the 7th and 8th data.
So:
Q_3 = \frac{1}{2}(12+15)Q
3
=
2
1
(12+15)
Q_3 = \frac{1}{2}*27Q
3
=
2
1
∗27
Q_3 = 13.5Q
3
=13.5
Answer:
s = 37.81 m
Explanation:
Given that,
Mass of an apple, m = 1.5 kg
The apple falls for 2.75s before striking the ground.
We need to find how far the apple dropped before hitting the ground. It means we need to find the distance covered. Using the second equation of motion to find it :
u is initial velocity, u = 0 (at rest)
So, Apple dropped 37.81 m before hitting the ground.
Answer:
0.50m/s
Explanation:
Average velocity is the change in displacement of a body with respect to time.
Velocity = ∆S/∆t
∆S = 100m - 70m
∆S = 30m
∆t = 2min - 1 min
∆t = 1min = 60secs
Substitute the given parameters into the formula for velocity
Velocity = 30m/60s
Velocity = 1/2 m/s
Average Velocity = 0.5m/s
We know, F = k * q₁ * q₂ / r²
Substitute the known values,
F = 9 * 10⁹ * 5 * 7 / (1.2)²
F = 315 * 10⁹ / 1.44
F = 218.75 * 10⁹ N
F = 2.1875 * 10¹¹ N [ Final Answer ]
Hope this helps!
