33=17x-18 add 18 both sides
51=17x. divide 17 both sides
3 = x
Answer:

Step-by-step explanation:
we know that
A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form
or 
we have the points (2,14) and (4,28)
Find out the constant of proportionality k
For (2,14) ----> 
For (4,28) ----> 
so
the value of
therefore
The equation is
Answer:

Step-by-step explanation:
Let the two numbers be x and y
So we have:

using the elimination method, add the two equations together:

solve for x:

substitute our value for x back into the first equation we made:

therefore,

Answer:
see explanation
Step-by-step explanation:
Given
f(x) = ax + b
x = 2 → f(x) = 1 and x = - 1 → f(x) = - 5 ( from the table )
Substitute these values into f(x) = ax + b, that is
2a + b = 1 → (1)
- a + b = - 5 → (2)
Subtract (2) from (1) term by term to eliminate b
3a = 6 ( divide both sides by 3 )
a = 2
Substitute a = 2 into (2) and evaluate for b
- 2 + b = - 5 ( add 2 to both sides )
b = - 3
(b)
When x maps onto itself then
ax + b = x, that is
2x - 3 = x ( subtract x from both sides )
x - 3 = 0 ( add 3 to both sides )
x = 3
Thus a = 2, b = - 3 and x = 3
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.