ABCD is a parallelogram. AB>AD. Prove: m∠ADB>m∠BDC. Show your work (statement reason maybe?)
2 answers:
In the parallelogram ABCD, join BD.
Consider the triangle Δ ABD.
It is given that AB > AD.
Since, in a triangle, angle opposite to longer side is larger, we have,
∠ ADB > ∠ ABD. --- (1)
Also, AB || DC and BD is a transversal.
Therefore,
∠ ABD = ∠ BDC
Substitute in (1), we get,
∠ ADB > ∠ BDC.
Answer:
Step-by-step explanation:
Given that ABCD is a parallelogram and AB >AD
To prove that
let us join BD.
Consider triangle ABD. AB and AD are two sides that AB>AD
By triangle theorem we get
Since opposite sides are parallel, we find that AB and Dc are parallel lines with BD as transversal.
Hence (alternate angles)
It follow that
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Answer:
0.048 is the probability that more than 950 message arrive in one minute.
Step-by-step explanation:
We are given the following information in the question:
The number of messages arriving at a multiplexer is a Poisson random variable with mean 15 messages/second.
Let X be the number of messages arriving at a multiplexer.
Mean = 15
For poison distribution,
Mean = Variance = 15
From central limit theorem, we have:
where n is the sample size.
Here, n = 1 minute = 60 seconds
P(x > 950)
Calculation the value from standard normal z table, we have,
0.048 is the probability that more than 950 message arrive in one minute.