You add 1/4 each year:
800 to 1000 in one year
1000 to 1250 in 2
1250 to around 1812.5
Answer:
x = -36, y = 6, z = -6
Step-by-step explanation:
The requirement x/z = -z means x = -z².
The requirement x/y = z means x = yz.
These two requirements together mean yz = -z², or y = -z.
The requirements that z/2 and z/3 are integers mean that z is a multiple of 2·3 = 6. The smallest magnitude non-zero multiple is z=-6 (since we also require z < -z).
Using z=-6, we have x = -z² = -36; y = -(-6) = 6.
For some positive integer n, ...
... x = -36n², y = 6n, z = -6n.
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>
Answer:
Well 14 2/3 is 14.66 so it would be B but yea it would fit.
Step-by-step explanation:
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