We use the equation for repeated trials written below:
Probability = n!/r!(n-r)! * p^(n-r) * q^r
The p is the probability of getting a side A in one toss. Since a counter has only two side, p = 0.5. The q is the probability of not getting side A in one toss, which is also q = 0.5. Now, r is the number of success per n trials. There are 3 tosses so, n=3. The question is getting "at least 1" counter. So, r=1, r=2 and r=3.
Probability for r=1: 3!/1!(3-1)! * (0.5)^(3-1) * (0.5)^1= 0.375
Probability for r=2: 3!/2!(3-2)! * (0.5)^(3-2) * (0.5)^2= 0.375
Probability for r=1: 3!/3!(3-3)! * (0.5)^(3-3) * (0.5)^3= 0.125
Total probability = 0.375 + 0.375 + 0.125 = 0.875
It is d, you first have to see how many times 0.2 goes into 18
Step-by-step explanation:
45X + 2000 = Y
Where X = Number of units
Y = Total Cost
Answer:
n=5
Step-by-step explanation:
6(n-2)-8=22+4(2-n)
open brackets
6n-12-8=22+8-4n
collect like terms
6n+4n=22+8+12+8
10n=50
n=50/10
n=5
