Answer:
The answer is below
Step-by-step explanation:
Let S denote syntax errors and L denote logic errors.
Given that P(S) = 36% = 0.36, P(L) = 47% = 0.47, P(S ∪ L) = 56% = 0.56
a) The probability a program contains both error types = P(S ∩ L)
The probability that the programs contains only syntax error = P(S ∩ L') = P(S ∪ L) - P(L) = 56% - 47% = 9%
The probability that the programs contains only logic error = P(S' ∩ L) = P(S ∪ L) - P(S) = 56% - 36% = 20%
P(S ∩ L) = P(S ∪ L) - [P(S ∩ L') + P(S' ∩ L)] =56% - (9% + 20%) = 56% - 29% = 27%
b) Probability a program contains neither error type= P(S ∪ L)' = 1 - P(S ∪ L) = 1 - 0.56 = 0.44
c) The probability a program has logic errors, but not syntax errors = P(S' ∩ L) = P(S ∪ L) - P(S) = 56% - 36% = 20%
d) The probability a program either has no syntax errors or has no logic errors = P(S ∪ L)' = 1 - P(S ∪ L) = 1 - 0.56 = 0.44
B because k is not a specified number
ANSWER
EXPLANATION
The diagram represents the net of a rectangular prism.
The dimensions are,
and
The total surface area of a rectangular prism is given by:
We substitute the values to obtain,
Range: maximum - minimum
IQR: third quartile - first quartile
median: middle number once all numbers are put in numerical order
<h2>>>> Answer <<<</h2>
Let's check which polynomial is divisible by ( x - 1 ) using hit , trial and error method .
A ( x ) = 3x³ + 2x² - x
The word " divisible " itself says that " it is a factor "
Using factor theorem ;
Let;
=> x - 1 = 0
=> x = 1
Substitute the value of x in p ( x )
p ( 1 ) =
3 ( 1 )³ + 2 ( 1 )² - 1
3 ( 1 ) + 2 ( 1 ) - 1
3 + 2 - 1
5 - 1
4
This implies ;
A ( x ) is not divisible by ( x - 1 )
Similarly,
B ( x ) = 5x³ - 4x² - x
B ( 1 ) =
5 ( 1 )³ - 4 ( 1 )² - 1
5 ( 1 ) - 4 ( 1 ) - 1
5 - 4 - 1
5 - 5
0
This implies ;
B ( x ) is divisible by ( x - 1 )
Similarly,
C ( x ) = 2x³ - 3x² + 2x - 1
C ( 1 ) =
2 ( 1 )³ - 3 ( 1 )² + 2 ( 1 ) - 1
2 ( 1 ) - 3 ( 1 ) + 2 - 1
2 - 3 + 2 - 1
4 - 4
0
This implies ;
C ( x ) is divisible by ( x - 1 )
Similarly,
D ( x ) = x³ + 2x² + 3x + 2
D ( 1 ) =
( 1 )³ + 2 ( 1 )² + 3 ( 1 ) + 2
1 + 2 + 3 + 2
8
This implies ;
D ( x ) is not divisible by ( x - 1 )
<h2>Therefore ; </h2>
<h3>B ( x ) & C ( x ) are divisible by ( x - 1 ) </h3>
