# Waec Further Maths answers 2023

F/MATHS-OBJ
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MORE COMING

2. Let V be the volume of the cube and s be the length of a side. We know that V = s^3, so differentiating both sides with respect to time t gives:

dV/dt = 3s^2 ds/dt

We are given that dV/dt = 3 cm^3/s, and s = 6 cm. Substituting these values into the equation above, we get:

3 = 3(6)^2 ds/dt

Solving for ds/dt, we get:

ds/dt = 1/12 cm/s

Therefore, the rate of change of the side of the base when its length is 6 cm is 1/12 cm/s.

(5)
TABULATE

Mass (kg): 31-40| 41-50| 51-60| 61-70| 71-80| 81-90

F: 3| 10| 15| 12| 6| 4

x: 35.5| 45.5| 55.5| 65.5| 75.5| 85.5

Fx: 106.5| 455| 832.5| 786| 453| 342

Class boundaries: 30.5-40.5| 40.5-50.5| 59.5-60.5| 60.5-70.5| 70.5-80.5| 80.5-90.5

(i) Mean (x-bar) = ∑fx/∑f = 2975/50 = 59.5
:. Mean = 60kg

(ii) Mode = L¹ + ( fm-fa/2fm-fa-fb) c
= 50.5. + (15-10/2*15-10-12)10
= 50.5 +(5/8)10
= 50.5+6.25
:. Mode = 56.75
Mode = 57kg(approx)

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