Answer:
B
Step-by-step explanation:
The amortization period in months is:
30 years * 12 = 360 months
The monthly interest rate would be 5.25%/12 = 0.004375
The payment for monthly mortgage formula would be:

Where
E is the monthly mortgage payment
C is the cost of mortgage, cost is $150,000
r is the monthly rate of interest, which is 0.004375
n is the period, in months, which is 360
Substituting, we get our answer:

So, the correct answer is B
Answer:
Parallel
Step-by-step explanation:
These 2 lines are parallel because they have the same slope (7).
Answer:
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Step-by-step explanation:
These are some quotations that I don’t know
Answer: Hence, the probability that he will get at least one lemon is 0.70.
Step-by-step explanation:
Since we have given that
Number of cars = 30
Number of lemon cars = 10
Number of other than lemon cars = 30-10 = 20
According to question, he bought 3 cars,
we need to find the probability that you will get at least one lemon.
So, P(X≤1)=1-P(X=0)=1-P(no lemon)
Here, P(no lemon ) is given by

so, it becomes,

Hence, the probability that he will get at least one lemon is 0.70.