Answer:
ii. 75 steps
iii. 75 steps
iv. 106 steps, and 
Step-by-step explanation:
Let Musah's starting point be A, his waiting point after taking 50 steps northward and 25 steps westward be B, and his stopping point be C.
ii. From the second attachment, Musah's distance due west from A to C (AD) can be determined as;
bearing at B =
, therefore <BCD = 
To determine distance AB,
=
+ 
= 25000 + 625
= 3125
AB = 
= 55.90
AB ≅ 56 steps
Thus, AC = 50 steps + 56 steps
= 106 steps
From ΔACD,
Sin
= 
⇒ x = 106 × Sin 
= 74.9533
≅ 75 steps
Musah's distance west from centre to final point is 75 steps
iii. From the secon attachment, Musah's distance north, y, can be determined by;
Cos
= 
⇒ y = 106 × Cos 
= 74.9533
≅ 75 steps
Musah's distance north from centre to final point is 75 steps.
iv. Musah's distance from centre to final point is AC = AB + BC
= 50 steps + 56 steps
= 106 steps
From ΔACD,
Tan θ = 
= 1.0
θ =
1.0
= 
Musah's bearing from centre to final point =
+ 
= 
Answer:
12 p - 8 p, 4p ( 3 - 2)
Step-by-step explanation:
p = equal the number of ounces of pasta salad in one container
then 12 × p = 12 p in 12 containers
the students finished the pasta salad in 8 containers which equals = 8 p
the number of ounces of pasta left = 12 p - 8 p
b) using the distributive property for example a ( b + c) = ( a×b) + (a ×c)
12 p - 8 p = 4p ( 3 - 2)
A) 40 out of 100 are red.
The probability would be 40/100, which reduces to 2/5
B) 40 out of the 100 are purple.
The probability would be 40/100, which reduces to 2/5
C) the two probabilities are the same.