Answer:
(0,2)
Step-by-step explanation:
solve by addition/elimination
2x + 3y= 6
–3x + 5y = 10
multiply first equation by 3 and second one by 2 to eliminate x)
6x+9y=18
-6x+10y=20 (add the two equations)
6x+9y-6x+10y=38
19y=38
y=38/19=2
2x+3y=6
2x=6-6
x=0
I think it should be angle bisector
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:
so im guessing the remainder of hw problems->
Step-by-step explanation:
24-7 =17 math problems left from doing some lunch
17-7= 10 math so its 10 math problems left
