I need help with no.7
1 answer:
Answer:
(a) 315°
(b) 3°
(c) 238°
Step-by-step explanation:
Bearings are measured clockwise from north. The triangle described is illustrated in the attachment.
<h3>(a)</h3>The bearing of P from R is 180° different from the bearing of R from P it will be ...
135° +180° = 315° . . . . bearing of P from R
__
<h3>(b)</h3>The bearing of Q from R is 48° more than the bearing of P from R, so is ...
315° +48° = 363°, or 3° . . . . bearing of Q from R
__
<h3>(c)</h3>The angle QPR has a value that makes the sum of angles in the triangle equal to 180°. It is ...
180° -48° -55° = 77°
The bearing of Q from P is 77° less than the bearing of R from P, so is ...
135° -77° = 58°
As above, the reverse bearing from Q to P is ...
58° +180° = 238° . . . . bearing of P from Q
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Step-by-step explanation:
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Have a nice day!
I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)
- Heather
Answer:
A continuous probability distribution having a rectangular shape, where the probability is evenly distributed over an interval of numbers is a(n) __uniform__________ distribution
Step-by-step explanation:
Given that there is a continuous probability distribution having a rectangular shape, where the probability is evenly distributed over an interval of numbers
Since the pdf is rectangular in shape and total probability is one we can say all values in the interval would be equally likely
Say if the interval is (a,b) P(X) = p the same for all places
Since total probability is 1,
we get integral of P(X)=p(b-a) =1
Or p=
this is nothing but a uniform distribution continuous defined in the interval
A continuous probability distribution having a rectangular shape, where the probability is evenly distributed over an interval of numbers is a(n) __uniform__________ distribution