i. 171
ii. 162
iii. 297
Solution,
n(U)= 630
n(I)= 333
n(T)= 168
i. Let n(I intersection T ) be X
<h3>ii.
n(only I)= n(I) - n(I intersection T)</h3><h3>
= 333 - 171</h3><h3>
= 162</h3>
<h3>
iii. n ( only T)= n( T) - n( I intersection T)</h3><h3>
= 468 - 171</h3><h3>
= 297</h3>
<h3>
Venn- diagram is shown in the attached picture.</h3>
Hope this helps...
Good luck on your assignment...
Answer:
the required probability is 0.09
Option a) 0.09 is the correct Answer.
Step-by-step explanation:
Given that;
mean μ = 7
x = 4
the probability of exactly 4 bridge construction projects taking place at one time in this state = ?
Using the Poisson probability formula;
P( X=x ) = ( e^-μ × u^x) / x!
we substitute
P(X = 4) = (e⁻⁷ × 7⁴) / 4!
= 2.1894 / 24
= 0.0875 ≈ 0.09
Therefore the required probability is 0.09
Option a) 0.09 is the correct Answer.
It is the graph shifted 6 units right
Area is 50 m^2
5 x 3 = 15
7 x 5 = 35
35 + 15 = 50 m^2
Answer:
the first one is the answer
