<span>When order does not matter you use what is called n choose k formula
n!/(k!(n-k)!) where n is the number of things to choose from and k is the number of choices made, in this case:
20!/(4!(20-4)!)=4845</span>
Answer:
{1, 5, 25, 125, 625}
Step-by-step explanation:
The smallest positive integers that meet the requirement will be ...
5^0 = 1
5^1 = 5
5^2 = 25
5^3 = 125
5^4 = 625
As a set, these numbers are {1, 5, 25, 125, 625}.
Answer: t = 10
Step-by-step explanation:m
Given that; n₁ = 10, n₂ = 10
ж₁ = 50, ж₂ = 30
Sˣ₁ = 20, Sˣ₂ = 20
Now using TEST STATISTICS
t = (ж₁ - ж₂) / √ ( Sˣ₁/n₁ + Sˣ₂/n₂ )
so we substitute our figures
t = ( 50 - 30 ) / √ ( 20/10 + 20/10 )
t = 20 / √4
t = 10
Let one of the number be x and the other be y
<span>two times the first number plus the second number is 41
</span>2x + y = 41
<span>the first number plus three times the second number is 73
x + 3y = 73
</span>2x + y = 41 --------- (1)
x + 3y = 73 --------- (2)
From (1):
2x + y = 41
y = 41 - 2x ------- sub into (2)
x + 3(41-2x) = 73
x + 123 - 6x = 73
x -6x = 73 - 123
-5x = -50
x = 10 -------- sub into (1)
2 (10) + y = 41
20 + y = 41
y = 41 - 20
y = 21
Ans: x = 10, y = 21
