Answer:
0.04
Step-by-step explanation:
Given that a dart is to be thrown at a target. The probability the dart will hit the target (yes or no) on a single attempt is 0.20
Each throw is independent of the other throws.
Let x be the no of times the target is hit in the trial of 4 throws
Since each outcome is independent of the other X is binomial with n =4 and p = 0.20
Required probability
= the probability that it will take less than or equal to 4 throws to hit the target on both successful target hits
=P(X=2 when n =2) + P(x=2 when x =3) + P(x=2 when x =4)
= 
Answer:
Step-by-step explanation:
First, we gather the data:
The model is a single 
landing strip with 
This based on the assumption that service times are exponential.
Then, we must find the maximum rate such that 
Using the table, we find that:
if and only if
=
Sorry bro I don’t know the answer but try sharing it again because it is not showing for many people
Well, that would be 1.5x7.. which is 10.5 :) your welcome
Answer:
0.0351478382 (To be precise)
Step-by-step explanation: Can I get brainliest? Thanks
1. Normal Distribution --> Z ~ (0,1^2)
2. Use normalcdf(lower bound, upper bound, μ, σ) function on a graphing calculator
P(Z≥103.53) = normalcdf(103.53, 1e99 [default], 80, 13)
P(Z≥103.53) ≈ 0.03
3. μ+σ ≈ 13.59% According to Z-distribution chart
80+13=93
So about 93 exceed only the top 16% (estimated answer not exact)
