Answer:
The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.
Step-by-step explanation:
We can model this with a binomial random variable, with sample size n=20 and probability of success p=0.08.
The probability of k online retail orders that turn out to be fraudulent in the sample is:
We have to calculate the probability that 2 or more online retail orders that turn out to be fraudulent. This can be calculated as:
![P(x\geq2)=1-[P(x=0)+P(x=1)]\\\\\\P(x=0)=\dbinom{20}{0}\cdot0.08^{0}\cdot0.92^{20}=1\cdot1\cdot0.189=0.189\\\\\\P(x=1)=\dbinom{20}{1}\cdot0.08^{1}\cdot0.92^{19}=20\cdot0.08\cdot0.205=0.328\\\\\\\\P(x\geq2)=1-[0.189+0.328]\\\\P(x\geq2)=1-0.517=0.483](https://tex.z-dn.net/?f=P%28x%5Cgeq2%29%3D1-%5BP%28x%3D0%29%2BP%28x%3D1%29%5D%5C%5C%5C%5C%5C%5CP%28x%3D0%29%3D%5Cdbinom%7B20%7D%7B0%7D%5Ccdot0.08%5E%7B0%7D%5Ccdot0.92%5E%7B20%7D%3D1%5Ccdot1%5Ccdot0.189%3D0.189%5C%5C%5C%5C%5C%5CP%28x%3D1%29%3D%5Cdbinom%7B20%7D%7B1%7D%5Ccdot0.08%5E%7B1%7D%5Ccdot0.92%5E%7B19%7D%3D20%5Ccdot0.08%5Ccdot0.205%3D0.328%5C%5C%5C%5C%5C%5C%5C%5CP%28x%5Cgeq2%29%3D1-%5B0.189%2B0.328%5D%5C%5C%5C%5CP%28x%5Cgeq2%29%3D1-0.517%3D0.483)
The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.
Answer:
-12r² + 10xr - 15x + 34r - 24
Step-by-step explanation:
1. Organize it, variables first - as well as adding constants such as 4 and -7
(5x - 6r + 8) (2r - 3)
2. Start by multiplying the 5x by (2r - 3), then -6r, followed by 8
(10xr - 15x) + (-12r^2 + 18r) + (16r - 24)
3. Simplify
10xr - 15x - 12r² + 18r + 16r - 24
4. Order by greatest to smallest factorial
-12r² + 10xr - 15x + 18r + 16r - 24
5. Combine like variables
-12r² + 10xr - 15x + 34r - 24
Answer:
c. 18
Step-by-step explanation:
First, you would find the total sum of the 60 numbers. The average is always the sum of the numbers divided by the number of numbers.
35*18=630
15*10=150
Now we do not know how many numbers the last group has. However, we can find it out doing 60-35-15 to see how many numbers are left. There would be 10 numbers left
10*30=300
630+150+300=1080
1080/60=18
Hence, the answer is C(18)
