Below is the pseudocode of the "test" algorithm:
algorithm "test"
var
A: VECTOR [1..3,1..3] OF WHOLE
C, L, M: WHOLE
start
RANDOM 1,50
FOR L OF 1 ATE 3 KNIFE
FOR C OF 1 ATE 3 KNIFE
READ (A [L, C])
SE (L = 1) E (C = 1) ENTAO
M <-A [L, C]
END IF
IF A [L, C]> M ENTAO
M <-A [L, C]
END IF
FIMPARA
FIMPARA
WRITING ("**** MATRIX A ****")
FOR L OF 1 ATE 3 KNIFE
FOR C OF 1 ATE 3 KNIFE
WRITE (A [L, C]: 4)
FIMPARA
WRITING
FIMPARA
WRITING (M)
final algorithm
On this algorithm, analyze the following sentences:
I. A is a matrix whose elements are integers generated randomly.
II. A is an order 3 matrix.
III. At the end of its execution, the algorithm prints the smallest value of the matrix.
IV. At the end of its execution, the algorithm prints the largest value of the matrix.
Now, mark the alternative that presents the CORRECT answer.
Alternatives:
The)
Only affirmative III is correct.
B)
Affirmations I and II are correct.
w)
Affirmations I, II and III are correct.
(d)
Affirmations I, II and IV are correct.
and)
All statements are correct.
Answer:
nope sorry man
Step-by-step explanation:
Answer:
xxxxxxzzzjjznzjsbsbsbshj
Step-by-step explanation:
sekjxhxjdjdknxxbbxjdkekmdkdkddo
Answer:
16/33
Step-by-step explanation:
We have a bag of :
9 red marbles
6 blue marbles
7 green marbles
11 yellow marbles
Total number of marbles = 33 marbles.
The possibility that a red or green marble will be selected from a bag
P( Red or Green) = P(Red) + P(Green)
In the question we are not told if it is with replacement or without. We do both
With replacement
P( R or G) = P(Red) + P(Green)
P(Red)= 9/33
P(Green) = 7/33
= 9/33 + 7/33
= 16/33
Therefore, the possibility that a red or green marble will be selected from a bag is 16/33
Answer:
Step-by-step explanation:
take 45 degree as reference angle
using sin rule
sin 45=opposite/hypotenuse
1/
=x/26
x=26/
x=13
x=18.4
