Suppose the number of births that occur in a hospital can be assumed to have a Poisson distribution with parameter = the average
1 answer:
Answer:
0.5372
Step-by-step explanation:
Given that the number of births that occur in a hospital can be assumed to have a Poisson distribution with parameter = the average birth rate of 1.8 births per hour.
Let X be the no of births in the hospital per hour
X is Poisson
with mean = 1.8
the probability of observing at least two births in a given hour at the hospital
=
the probability of observing at least two births in a given hour at the hospital = 0.5372
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Answer:
Step-by-step explanation:
NA = √[(- 4 - 1 )² + (- 3 - 2)²] = 5√2
AT = √[(8 - 1 )² + (1 - 2)²] = 5√2
TS = √[(3 - 8 )² + (- 4 - 1)²] = 5√2
NS = √[(- 4 - 3 )² + (- 3 + 4)²] = 5√2
NA = AT = TS = NS = 5√2
= (- 3 - 2) / (- 4 - 1) = 1 ........ <em>(1)</em>
= (- 4 - 1) / (3 - 8 ) = 1 ......... <em>(2)</em>
From (1) and (2) ⇒ NA║TS
= ( 1 - 2) / ( 8 - 1) = - 1 / 7 .......... <em>(3)</em>
= ( - 4 + 3) / ( 3 + 4) = - 1 / 7 .... <em>(4)</em>
From (3) and (4) ⇒ AT║NS
Thus, NATS is rhombus.