The answer is d (i.e. 51).
the first 9 pages utilize 9 digits. the remaining 21 pages utilize two digits number. Therefore,
21*2=42
the total is 42+9=51
Answer:
All possible are:
(G,L,S)
(G,L,R)
(G,L,P)
(G,S,R)
(G,S,P)
(G,R,P)
(L,S,R)
(L,S,P)
(S,R,P)
{L,R,P)
Probability of 1st/2nd/10th sample = 1/10
Step-by-step explanation:
All the possible combinations of the 3 size samples from a 5 size population have been listed without repetition.
Total Numbers of Samples = 10
To find the probability of finding the first sample from random sampling procedure,
Probability = Number of desired outcomes/ Total number of outcomes
Where Number of desired outcome is 1 and total number of outcomes is 10.
Probability = 1/10
Similarly, to find 2nd sample or 10th sample, the number of desired outcomes is same i.e 1, hecne the probability remains the same i.e 1/10
39. because its close to it lol.
9.58 = 9 58/100 = 9 29/50
39x-3y=21
-3y=21-39x
(subtract both sides of the equation by 39x and divide both sides by -3)
y=-7+3x : would be the equation for y.
