What is the equation of a line that is perpendicular to the line y= -1/4x - 1 and passes through the point (3,7) ?
1 answer:
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Answer:
Step-by-step explanation:
We know that probability for any event =
Given : Charlotte has 6 cherry candies, 3 grape candies, and 3 lime candies.
I..e Total pieces of candies she has = 6+3+3= 12
Now , If Charlotte randomly pulls one piece of candy out of the bag, what is the probability that it will be cherry is given by :-
Hence, the probability that it will be cherry is .
Answer:
See answer below
Step-by-step explanation:
Hi there,
The volume of a cube is the edge length cubed:
³
The edge length given is 3 u, so plug into the formula:
³
³
thanks,
200 per month for 5 years = 12,000
Phil can (at most) pay 12,000 for the car loan.
That would be the payment total.
So, how do we solve for the PRINCIPAL?
Looking at the attached formula, we see the
principal = Total - Total * ((1 +r)^-n) / r * n
where n = # of payments = 5 year * 12 months = 60 and
rate / 1.200 = 7.5/1,200 = .00625
principal = [12,000 - 12,000 * (1.00625)^-60] / .375
principal = [12,000 - 12,000 * <span> <span> <span> 0.6880918239 </span> </span> </span> ] / .375
principal = [12,000 - <span> <span> <span> 8,257.1018863488] / .375
</span></span></span>principal = <span> <span> <span> 3,742.</span></span></span>90 / .375
principal = 9,981.06
Source:
http://www.1728.org/loanfrm4.htm
Phil can (at most) pay 12,000 for the car loan.
That would be the payment total.
So, how do we solve for the PRINCIPAL?
Looking at the attached formula, we see the
principal = Total - Total * ((1 +r)^-n) / r * n
where n = # of payments = 5 year * 12 months = 60 and
rate / 1.200 = 7.5/1,200 = .00625
principal = [12,000 - 12,000 * (1.00625)^-60] / .375
principal = [12,000 - 12,000 * <span> <span> <span> 0.6880918239 </span> </span> </span> ] / .375
principal = [12,000 - <span> <span> <span> 8,257.1018863488] / .375
</span></span></span>principal = <span> <span> <span> 3,742.</span></span></span>90 / .375
principal = 9,981.06
Source:
http://www.1728.org/loanfrm4.htm