divide total miles by time:
157.5 / 2.5 = 63 miles per hour
It's c. all of them added up is 41/24 which converts to 1 17/24.
Answer:
The sample size n = 4225
Step-by-step explanation:
We will use maximum error formula = 
but we will find sample size "n"

Squaring on both sides , we get

Given 99% confidence interval (z value) = 2.56
given maximum error = 0.02

n≤
( here S.D = p(1-p) ≤ 1/2
on simplification , we get n = 4225
<u>Conclusion</u>:
The sample size of two samples is n = 4225
<u>verification</u>:-
We will use maximum error formula =
=
= 0.0196
substitute all values and simplify we get maximum error is 0.02
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:
dial it up
2:16
simplify
1:8
now multiply 8 by the number of gigs
2:16
3:24
and so on.....
ring ring
1:9
again multiply the cost by the number of gigs
2:18
3:27
and so on....