Easiest way is if you substitute each point (x,y) into each set of equations and both points work for both equations in the system of equations, then it is the correct answer
Otherwise substitute one equation for y in the other equation:
2x + 6 = x^2 + 5x + 6
-2x - 6. -2x -6
0 = x^2 + 3x. Factor
0 = x (x + 3)
Solve: x = 0. x + 3 = 0. ——> x = -3. Substitute into one original equation to get y value for
y = 2x + 6.
y = 2(0) + 6. y = 2(-3) + 6
y = 6. y = -6 + 6 —-> y = 0
(0 , 6) And. (-3 , 0)
Answer:
V= 8/3
Step-by-step explanation:
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>
