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tiny-mole [99]
1 year ago
8

If ​P(A)=0.9​, ​P(B)​=0.1, and A and B are​ independent, find​ P(A and​ B).

Mathematics
1 answer:
sweet [91]1 year ago
6 0

According to the given statement;

If ​P(A)=0.9​, ​P(B)​=0.1, and A and B are​ independent, ​ P(A and​ B)=0.09

<h3>What is probability of event?</h3>

The area of mathematics known as probability deals with numerical representations of the likelihood that an event will occur or that a proposition is true. An event's probability is a number between 0 and 1, where, roughly speaking, 0 denotes the event's impossibility and 1 denotes certainty. The likelihood that an event will occur increases with its probability. A straightforward illustration is tossing a fair (impartial) coin. Since there are no other possible outcomes and the coin is fair, the odds of both the outcomes, "heads" and "tails," are equally likely to occur. As a result, the probability of either outcome is half.

<u>According to the given value;</u>

P(A) = 0.9

P(B) = 0.1

A and B independent;

We know that independent

P(A∩B) = P(A) · P(B)

             = (0.9)*(0.1)

             =0.09

P(A∩B) =0.09

hence, If ​P(A)=0.9​, ​P(B)​=0.1, and A and B are​ independent, ​ P(A and​ B)=0.09

To more about probability of event, visit

brainly.com/question/17089724

#SPJ9

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Answer:

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Step-by-step explanation:

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Speed = Distance ÷ Time ⇔ Distance = Speed × Time

→ Convert 1 hour and 47 minutes into decimal format

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3 years ago
An aeroplane X whose average speed is 50°km/hr leaves kano airport at 7.00am and travels for 2 hours on a bearing 050°. It then
Zigmanuir [339]

Answer:

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Step-by-step explanation:

Plane X with an average speed of 50km/hr travels for 2 hours from P (Kano Airport) to point Q in the diagram.

Distance = Speed X Time

Therefore: PQ =50km/hr X 2 hr =100 km

It moves from Point Q at 9.00 am and arrives at the airstrip A by 11.30am.

Distance, QA=50km/hr X 2.5 hr =125 km

Using alternate angles in the diagram:

\angle Q=110^\circ

(a)First, we calculate the distance traveled, PA by plane Y.

Using Cosine rule

q^2=p^2+a^2-2pa\cos Q\\q^2=100^2+125^2-2(100)(125)\cos 110^\circ\\q^2=34175.50\\q=184.87$ km

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Time taken =1.5 hour

Therefore:

Average Speed of Y

=184.87 \div 1.5\\=123.25$ km/hr\\\approx 123$ km/hr (correct to three significant figures)

(b)Flight Direction of Y

Using Law of Sines

\dfrac{p}{\sin P} =\dfrac{q}{\sin Q}\\\dfrac{125}{\sin P} =\dfrac{184.87}{\sin 110}\\123 \times \sin P=125 \times \sin 110\\\sin P=(125 \times \sin 110) \div 184.87\\P=\arcsin [(125 \times \sin 110) \div 184.87]\\P=39^\circ $ (to the nearest degree)

The direction of flight Y to the nearest degree is 39 degrees.

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