B) 14.
If you start with subtracting the 8 pages from 50, you get 42 remaining pages. Then it says 2/3 of the REMAINING pages so you divide 42 into thirds. 42/3=14. You want 2/3 of that, so you multiply the 14 by two to get 28 (it's like there are three piles of 14 pages and you want only 2 out of the three piles). So now you've used 28 pages + the original 8. You've used a TOTAL of 36 pages. Subtract 36 from 50 and you get 14 left.
Answer:
X=2
Step-by-step explanation:
Answer:
Step-by-step explanation:
<u>Equation Solving</u>
We are given the equation:
It's required to solve it for a.
Swap sides to have the letter on the left side:
Divide by 
:
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>
