1.)
<span>((i <= n) && (a[i] == 0)) || (((i >= n) && (a[i-1] == 0))) </span>
<span>The expression will be true IF the first part is true, or if the first part is false and the second part is true. This is because || uses "short circuit" evaluation. If the first term is true, then the second term is *never even evaluated*. </span>
<span>For || the expression is true if *either* part is true, and for && the expression is true only if *both* parts are true. </span>
<span>a.) (i <= n) || (i >= n) </span>
<span>This means that either, or both, of these terms is true. This isn't sufficient to make the original term true. </span>
<span>b.) (a[i] == 0) && (a[i-1] == 0) </span>
<span>This means that both of these terms are true. We substitute. </span>
<span>((i <= n) && true) || (((i >= n) && true)) </span>
<span>Remember that && is true only if both parts are true. So if you have x && true, then the truth depends entirely on x. Thus x && true is the same as just x. The above predicate reduces to: </span>
<span>(i <= n) || (i >= n) </span>
<span>This is clearly always true. </span>
Answer:
endl
Explanation:
Note that endl must be free of quotation marks; otherwise, the program will treat it as a string.. The \n Character. The other way to break a line in C++ is to use the newline character
Answer:
i think it helps
Explanation:
sorry there was someone else rude as heck messaging and responding I couldn't see their screen name
what isig naame
oh okay
I have sent dm
i dont know your name there
In our bag, 1/2 is peanuts, 1/4 is chocolate and 1/4 is dried fruit.
The likelihood of drawing a chocolate therefore is 1/4.
The likelihood of drawing a peanut is 1/2 and the likelihood of drawing a dried fruit is 1/4.
Thus, D is the correct answer because the 1/4 likelihood of drawing a chocolate is less than the 1/2 chance of drawing a peanut.