Consider this option:
C³₂₇=27!/(3!*24!)=25*13*9=2925 ways to select 3 students.
First we need to find out what kind
of logarithm rule is given, the given is logarithm product rule which states
that a log of a product is equal to the sum of the log of the first base and
the log of the second base.
By:
= log (1.37 x 10⁹) =
log (1.37) + log (10⁹)
= log (1.37) + 9
= 9 + log (1.37)
In the meantime, 1.37 is between
1 and 10 its logarithm will be between 0 and 1. Thus, the value of log (1.37 x 10⁹)
falls between 9 and 10 because when you compose a scientific notation you will
always have a number among 1 and 10 by 10 to some power. That power tells you
the integer part of the logarithm.
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Answer:
76% of X= 9.12. X= 12 gram. Therefore weight of the sample is 12 gram.
Step-by-step explanation:
i hope it's help you
Answer:
Step-by-step explanation:
We have been given that there are 4 red marbles and 7 blue marbles and 5 yellow marbles in a bag. Monica will randomly pick two marbles out of the bag replacing the first marble before picking the second marble.
Since Monica will replace the first marble before picking the second marble, therefore, probability of both events will be independent and probability of occurring one event will not affect the probability of second event's occurring.
Since the probability of two independent compound events is always the product of probabilities of both events.
Now let us find probability of picking a red marble out of 16 (4+7+5) marbles.
Probability of picking blue ball out of 16 (4+7+5) marbles:
Now let us find probability of Monica picking a red and then a blue marble.
Therefore, the probability of picking a red and then blue marble is .
Answer:
The probability that a randomly chosen code starts with M and ends with E is 0.05 ....
Step-by-step explanation:
According to the given statement we have to make five letter code from A, F, E, R, and M without repeating any letter. We have to find that what is probability that a randomly chosen code starts with M and ends with E.
Thus the probability of picking the first letter M = 1/5
After that we require the sequence (not E, not E, not E) which is equal to:
= 3/4 * 2/3* *1/2
= 1/4
Now multiply 1/5 and 1/4
1/5 * 1/4
= 1/20
= 0.05
Therefore the probability that a randomly chosen code starts with M and ends with E is 0.05 ....
