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.
A=x
B= 2x
C=3x
If you add them up =
6x =180
÷6 both sides
x=30
A=30
B=60
C=90
one angle is 90 degrees so the triangle is classified as a right triangle
Let us say Karen received X amount of money. Then Natasha has received (X+210).
The ratio that the money divided was 2:5.
X/(X+210) = 2/5
5X = 2(X+210)
3X = 420
X = 140
Total money shared
= X+(X+210)
= 140+140+210
= 490
So the total money shared was $490.
Point-slope form :-
... y = mx + b
<em>In this equation, 'm' represents the slope and 'b' represents the y-intercept. </em>
Given :-
... y = -1/7x + 2
<u><em>Hence, the slope is -1/7 and y-intercept is 2 or ( 0 , 2 ) </em></u>
Hope it helps!!
