The hypotenuse function is correct, but the main function has a problem. Explain why it may not work, and show a way to fix it.
1 answer:
Answer: hello your question is incomplete attached below is the missing codes
Answer : It may not work because; the double pointer P was not allotted a double using a new double ( I.e. Error in the main function ) also the cout statement was not written in the main function
Step-by-step explanation:
It may not work because; the double pointer P was not allotted a double using a new double ( I.e. Error in the main function ) also the cout statement should be moved to the Main
<u><em>The correction is below </em></u>
{
double* p = new double;
hypotenuse(1.5, 2.0, p);
cout << "The hypotenuse is " << *p << endl;
}
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Answer:
(a) 0.4096
(b) 0.64
(c) 0.7942
Step-by-step explanation:
The probability that the player wins is,
Then the probability that the player losses is,
The player is playing the video game with 4 different opponents.
It is provided that when the player is defeated by an opponent the game ends.
All the possible ways the player can win is: {L, WL, WWL, WWWL and WWWW)
(a)
The results from all the 4 opponents are independent, i.e. the result of a game played with one opponent is unaffected by the result of the game played with another opponent.
The probability that the player defeats all four opponents in a game is,
P (Player defeats all 4 opponents) =
Thus, the probability that the player defeats all four opponents in a game is 0.4096.
(b)
The probability that the player defeats at least two opponents in a game is,
P (Player defeats at least 2) = 1 - P (Player losses the 1st game) - P (Player losses the 2nd game) =
Thus, the probability that the player defeats at least two opponents in a game is 0.64.
(c)
Let <em>X</em> = number of times the player defeats all 4 opponents.
The probability that the player defeats all four opponents in a game is,
P(WWWW) = 0.4096.
Then the random variable
The probability distribution of binomial is:
The probability that the player defeats all the 4 opponents at least once is,
P (<em>X</em> ≥ 1) = 1 - P (<em>X</em> < 1)
= 1 - P (<em>X</em> = 0)
Thus, the probability that the player defeats all the 4 opponents at least once is 0.7942.