If sin θ = - \(\frac{3}{5}\) and θ lies in the third quadrant, find cos θ
\(\frac{4}{5}\)
- \(\frac{5}{4}\)
\(\frac{5}{4}\)
- \(\frac{4}{5}\)
Correct answer is D
Where sin θ = \(\frac{opp}{hyp}\) → \(\frac{-3}{5}\)
opp = -3, hyp = 5
using pythagoras formula
hyp\(^2\) = adj\(^2\) + opp\(^2\)
adj\(^2\) = hyp\(^2\) - opp\(^2\)
adj\(^2\) = 5\(^2\) - 3\(^2\) → 25 - 9
adj\(^2\) = 16
adj = 4
cos θ = \(\frac{adj}{hyp}\) → \(\frac{4}{5}\)
In third quadrant: cos θ is negative → - \(\frac{4}{5}\)
If the mean of 2, 5, (x+1), (x+2), 7 and 9 is 6. Find the median
5.5
5
6.5
6
Correct answer is C
Firstly; solving for x
6 = \(\frac{2 + 5 + x+1 + x+2 + 7 + 9}{6}\)
cross multiply to have:
6 * 6 = 2 + 5 + x+1 + x+2 + 7 + 9
36 = 2x + 26
36 - 26 = 2x
10 = 2x
x = 5
Median = \(\frac{7+6}{2}\)
→ 6.5
If a fair coin is tossed twice, what is the probability of obtaining at least one head?
0.25
0.75
0.5
0.33
Correct answer is B
No explanation has been provided for this answer.
The shaded portion in the venn diagram above represents?
F - (E n F) - (G n F)
E' n F n G'
E u F u G
F
Correct answer is B
-The shaded/stripped portion in the diagram is set F only.
-This implies that sets E and G though present in the Universal set, are not included in the expression.
-Mathematically; we have E' n F n G'
Solve for k in the equation \(\frac{1}{8}^{k+2}\) = 1
2
-4
-2
4
Correct answer is C
\((8^{-1})^{k+2}\) = \((8^{0})\)
base 8 cancel out on both sides
-1(k+2) = 0
-k -2 = 0
: k = -2
JAMB Subjects
Aptitude Tests