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Ksenya-84 [330]
3 years ago
10

A place from this table is chosen at random. Let event A = The place is a city.

Mathematics
1 answer:
Crank3 years ago
6 0

Answer:

Final answer is P(A^c)=\frac{3}{7}

Step-by-step explanation:

We have been given a table containing a list of few places that are either city or in North America.

Total number of places in that list = 7

That means sample space has 7 possible events.

Given that a place from this table is chosen at random. Let event A = The place is a city.

Now we need to find about what is P(A^c).

That means find find the probability that chosen place is not a city.

there are 3 places in the list which are not city.

Hence favorable number of events = 3

Then required probability is given by favorable/total events.

P(A^c)=\frac{3}{7}

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we multiply the lenght by the number:

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3 0
3 years ago
If tanA=a <br>then find sin4A-2sin2A/ sin4A+2sin2A​
anygoal [31]

Answer:

The value of the given expression is

\frac{sin4A-2sin2A}{sin4A+2sin2A}=-a^2

Step by step Explanation:

Given that tanA=a

To find the value of \frac{sin4A-2sin2A}{sin4A+2sin2A}

Let us find the value of the expression :

\frac{sin4A-2sin2A}{sin4A+2sin2A}=\frac{2cos2Asin2A-2sin2A}{2cos2Asin2A+2sin2A} ( by using the formula sin2A=2cosAsinA here A=2A)  

=\frac{2sin2A(cos2A-1)}{2sin2A(cos2A+1)}

=\frac{(cos2A-1)}{(cos2A+1)}

=\frac{(-(1-cos2A))}{(1+cos2A)}(using  sin^2A+cos^2A=1  here A=2A)

=\frac{-(sin^2A+cos^2A-(cos^2A-sin^2A))}{sin^2A+cos^2A+(cos^2A-sin^2A)}(using cos2A=cos^2A-sin^2A here A=2A)

=\frac{-(sin^2A+cos^2A-cos^2A+sin^2A)}{sin^2A+cos^2A+(cos^2A-sin^2A)}

=\frac{-(sin^2A+sin^2A)}{cos^2A+cos^2A}

=\frac{-2sin^2A}{2cos^2A}

=-\frac{sin^2A}{cos^2A}

=-tan^2A  ( using tanA=\frac{sinA}{cosA} here A=2A )

 =-a^2 (since tanA=a given )

Therefore \frac{sin4A-2sin2A}{sin4A+2sin2A}=-a^2

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3 years ago
If 35% of 300 children had zero cavities, how many children was that???
igomit [66]

Answer:

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Step-by-step explanation:

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105

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