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Jet001 [13]
4 years ago
12

Sheila bought a new computer for $2000 and has agreed to finance it at 12% interest, with $100 payments each month. When she mak

es her first payment next month, how much will she pay for interest alone?
Mathematics
1 answer:
mylen [45]4 years ago
5 0

Answer:

rate per month = .01 agreed = r

Luckily they only ask for the first month so it is 0.01 * 20 .

Later months get more complicated and it is like paying off a mortgage with the term unknown but the rate and payment known.

Payment = principal value [ r / {1-(1+r)^-n } ]

100 = 2000 [ .01 / {1 - (1.01)^-n } ]

5 = 1/ {1 - (1.01)^-n }

.2 = 1 - (1.01)^-n

.8 = 1.01^-n

log .8 = -n log 1.01

n = 22.4 months :)

so over the almost two years pay 2240 total :)

<em>* Hopefully this helps:) Mark me the brainliest:)!!!</em>

<em></em>

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A computer programmer has a 35 chance of finding a bug in any given program. what is the probability that she finds a bug within
vlada-n [284]

Answer:

Answer is d

Step-by-step explanation:

  • A computer programmer  has a 35 change of finding a bug in any given program.
  • we have to find probability that she finds a bug within the first 3 programs she examines
  • success = 0.35, failure = 0.65
  • She get a bug in the first program itself
  • p = 0.35
  • She get a bug in 2nd program
  • q = 0.65×0.35 =0.2275
  • She get a bug in the 3rd program
  • r = 0.65×0.65×0.35
  • r = 0.147875
  • Add all the three probabilities for answer
  • P = p+q+r
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7 0
3 years ago
Sum of 15 probability with two dice<br>​
Taya2010 [7]

Answer:

<h2>0</h2>

Step-by-step explanation:

all the outcomes you can get from rolling a dice twice are

2,3,4,5,6,7,8,9,10,11,12

15 is nowhere in those numbers so the probability of getting a sum of 15 is 0

6 0
3 years ago
If you had 1080 and the ratio is 3:4:5 how much would each person have
jolli1 [7]
270:360:450 is the ratio 
7 0
3 years ago
At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel
Alexeev081 [22]
Use Law of Cooling:
T(t) = (T_0 - T_A)e^{-kt} +T_A
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
42 = (80-20)e^{-3k} +20 \\  \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\  \\ k = -\frac{1}{3} ln (\frac{11}{30})

Substituting k back into cooling equation gives:
T(t) = 60(\frac{11}{30})^{t/3} + 20

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\  \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80
Solve for x:
x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80
Sub back into original cooling equation, x = T(t)
-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20
Solve for t:
60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3}  = 60 \\  \\  (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\  \\  \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\  \\ \\  t = 4.959

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM
8 0
3 years ago
Who is agent in marketing​
frozen [14]

Answer:

They act as an extension of the manufacturing company.

6 0
3 years ago
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