The number of ways or permutations to select five unordered elements from a set with three elements when repetition is allowed is 243.
According to the given question.
The total number of elements, n = 3
The number of elements to be selected at a time when repetition is allowed, r = 5
Since, we know that " the number of ways or permutations of n things taken r all at a time, when repetition of things are allowed, is
.
Therefore,
The number of ways or permutations to select five unordered elements from a set with three elements when repetition is allowed
= 
= 243
Hence, the number of ways or permutations to select five unordered elements from a set with three elements when repetition is allowed is 243.
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solution:
maximum acceleration produce by bag =60 g
time to come to stop = 36 ms
applying equation

inserting values


now to find distance of penctration:

inserting values


hence distance traveled by person before coming to rest is 0.38 m
Answer:
He gave you 48 cards.
Step-by-step explanation:
150 * 32% (which is also 0.32) is equal to 48.
Answer:
m = - 8 ± 6
Step-by-step explanation:
Given
m² + 16m - 8 = 0 ( add 8 to both sides )
m² + 16m = 8
To complete the square
add ( half the coefficient of the m- term )² to both sides
m² + 2(8)m + 64 = 8 + 64
(m + 8)² = 72 ( take the square root of both sides )
m + 8 = ±
= ±
= ± 6
Subtract 8 from both sides
m = - 8 ± 6