4x^2 = 12x
4x^2 - 12x = 0
4x(x - 3) = 0
4x = 0
x = 0...not this one
x - 3 = 0
x = 3 <== ur number
-4,5. 4,5 that would be the answer it's like the oppiste
Answer:
140° and 50°
Step-by-step explanation:
The supplement of the angle (180 - x)
The complement of the angle = (90 - x)
(180 -x) = 4(90-x) - 60
180 - x = 360 -4x - 60
180 -x = 300 - 4x
180 - x + 4x = 300
180 + 3x = 300
3x = 120
x = 40
The supplement (180 - x) = 180 - 40 = 140°
The complement (90 - x) = 90 - 40 = 50°
2/15 so 13.3% ?
I’m not entirely sure if this is right but I think it’s getting there...
Answer:
<em>The answers are for option (a) 0.2070 (b)0.3798 (c) 0.3938
</em>
Step-by-step explanation:
<em>Given:</em>
<em>Here Section 1 students = 20
</em>
<em>
Section 2 students = 30
</em>
<em>
Here there are 15 graded exam papers.
</em>
<em>
(a )Here Pr(10 are from second section) = ²⁰C₅ * ³⁰C₁₀/⁵⁰C₁₅= 0.2070
</em>
<em>
(b) Here if x is the number of students copies of section 2 out of 15 exam papers.
</em>
<em> here the distribution is hyper-geometric one, where N = 50, K = 30 ; n = 15
</em>
<em>Then,
</em>
<em>
Pr( x ≥ 10 ; 15; 30 ; 50) = 0.3798
</em>
<em>
(c) Here we have to find that at least 10 are from the same section that means if x ≥ 10 (at least 10 from section B) or x ≤ 5 (at least 10 from section 1)
</em>
<em>
so,
</em>
<em>
Pr(at least 10 of these are from the same section) = Pr(x ≤ 5 or x ≥ 10 ; 15 ; 30 ; 50) = Pr(x ≤ 5 ; 15 ; 30 ; 50) + Pr(x ≥ 10 ; 15 ; 30 ; 50) = 0.0140 + 0.3798 = 0.3938
</em>
<em>
Note : Here the given distribution is Hyper-geometric distribution
</em>
<em>
where f(x) = kCₓ)(N-K)C(n-x)/ NCK in that way all these above values can be calculated.</em>
