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3 years ago

Please help me with this problem...

Mathematics

1 answer:

Leokris [45]3 years ago

4 0

Answer:

The probability that they both miss their targets is 1/3

Step-by-step explanation:

In this question, we are trying to calculate the probability that if the captain and the pirate fires the canon in their ships at the same time, they both miss.

We proceed as follows;

Let the probability of the captain hitting the pirate be P(c) = 1/2

The probability of the captain missing is thus ; 1 - 1/2 = 1/2 or 0.5

Let the probability of the pirate hitting the captain be P(r) = 1/3 , the probability of him not hitting the captain is 1-1/3 = 2/3

Now the probability of both of them missing is = probability of the captain missing a fired shot * probability of the pirate missing a fired shot

Hence, we have; 2/3 * 1/2 = 2/6 = 1/3

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Answer:

Option b is the correct answer as both the equations are true for given solution.

Step-by-step explanation:

Given equations are:

-0.1x-0.3y=1.2

0.2x-0.5y=2

We can observe each graph and find the point that is the solution and put the point in the equations to know if that point is the solution

For option A:

(0,4)

Putting x=0 and y = 4 in both equations

$-0.1(0)-0.3(4)=1.2\\0-1.2 =1.2\\-1.2 \neq 1.2\\0.2(0)-0.5(4)=2\\0-2 = 2\\-2 \neq 2$

This is not the correct answer as both equations are not true with this solution

For Option B:

(0,-4)

Putting x = 0 and y = -4 in both equations

$-0.1(0)-0.3(-4)=1.2\\0+1.2 = 1.2\\1.2 = 1.2\\0.2x-0.5y=2\\0.2(0)-0.5(-4) = 2\\2 = 2$

Both equations are true for (0,-4) hence it is the solution of the system.

For Option C:

(4,0)

$-0.1(4)-0.3(0)=1.2\\-0.4-0 = 1.2\\-0.4 \neq 1.2\\0.2x-0.5y=2\\0.2(4)-0.5(0) = 2\\0.8 \neq 2$

Not true for both equations

Hence,

Option b is the correct answer.

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