Slope intercept = y=mx+b
m is the slope
b is the y int
point slope =y2-y1=m(x1-x2)
Number of students who have sent a text message only is given by the difference between those that have sent a text message and those that have done both.
i.e. Number of students who has sent a text message only = 58 - 12 = 46
Number of students who have uploaded a selfie only is given by the
difference between those that have uploaded a selfie and those that
have done both.
i.e. Number of students who have uploaded a selfie only = 21 - 12 = 9
The total number of students surveyed is 70.
Let the number of students who have neither sent a text message nor taken a selfie today be x, then
46 + 9 + 12 + x = 70
67 + x = 70
x = 70 - 67 = 3
Therefore, 3 students neither sent a text message nor taken a selfie today.
The number of students that have sent a text message or have taken a selfie today is given by the sum of the number of students that have sent a text message and the number of students that have taken a selfie less the number of people that have done both.
i.e. number of students that have sent a text message or have taken a selfie today = 58 + 21 - 12 = 67.
Answer:
CBA=MNP
Step-by-step explanation:
Side NP=CB
Side NM=AB
then the angles between those sides are equal
Well its
900+0+0
basically
like if it was
9,234
then it would be
9000+200+30+4
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>
