how can the mean of the primes of [ 1 2 3] be 2.5?
The mean of the primes which are 1 & 3 is 2.
1 is not a prime number, so therefore 2+3=5, 5/2=2.5
Test | Status | Code Input and Output |
---|---|---|
1 | Pass |
x = 3;
y_correct = 3;
assert(isequal(meanOfPrimes(x),y_correct))
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.]
[> In unix (line 32)
In meanOfPrimes (line 2)
In ScoringEngineTestPoint1 (line 3)
In solutionTest (line 3)]
|
2 | Pass |
x = [1 2 3];
y_correct = 2.5;
assert(isequal(meanOfPrimes(x),y_correct))
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.]
[> In unix (line 32)
In meanOfPrimes (line 2)
In ScoringEngineTestPoint2 (line 3)
In solutionTest (line 5)]
|
3 | Pass |
x = [3 3; 3 3];
y_correct = 3;
assert(isequal(meanOfPrimes(x),y_correct))
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.]
[> In unix (line 32)
In meanOfPrimes (line 2)
In ScoringEngineTestPoint3 (line 3)
In solutionTest (line 7)]
|
4 | Pass |
x = [7 3 8 8]';
y_correct = 5;
assert(isequal(meanOfPrimes(x),y_correct))
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.]
[> In unix (line 32)
In meanOfPrimes (line 2)
In ScoringEngineTestPoint4 (line 3)
In solutionTest (line 9)]
|
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