The amount Howie will pay back at the end of one year is $26000.00
The given parameters are:
-- the principal amount
-- the interest rate
--- the duration
The amount to pay back in this duration is then calculated using:
![A = P(1 + RT)](https://tex.z-dn.net/?f=A%20%3D%20P%281%20%2B%20RT%29)
So, we have:
![A = 25000(1 + 4\% \times 1)](https://tex.z-dn.net/?f=A%20%3D%2025000%281%20%2B%204%5C%25%20%5Ctimes%201%29)
![A = 25000(1 + 4\% )](https://tex.z-dn.net/?f=A%20%3D%2025000%281%20%2B%204%5C%25%20%29)
Express percentage as decimal
![A = 25000(1 + 0.04)](https://tex.z-dn.net/?f=A%20%3D%2025000%281%20%2B%200.04%29)
![A = 25000(1 .04)\\](https://tex.z-dn.net/?f=A%20%3D%2025000%281%20.04%29%5C%5C)
Multiply
![A = 26000](https://tex.z-dn.net/?f=A%20%3D%2026000)
Hence, the amount to pay back is $26000.00
Read more about simple interest at:
brainly.com/question/1115815
<span>√(5*8*24) or anything along those lines, you can switch the numbers around , it wont matter</span>
Answer:
The written form would be "Nine more than three times a number is greater than -18.
Step-by-step explanation:
To graph it, you must first solve for x.
3x + 9 > -18
3x > -27
x > -9
Now we can graph this by putting an open dot at -9 and then drawing an arrow to the right.
Answer:
a) P=0.0175
b) P=0.0189
Step-by-step explanation:
For both options we have to take into account that not only the chance of a "superevent" will disable both suppliers.
The other situation that will disable both is that both suppliers have their "unique-event" at the same time.
As they are, by definition, two independent events, we can calculate the probability of having both events at the same time as the product of both individual probabilities.
a) Then, the probability that both suppliers will be disrupted using option 1 is
![P_1=P_{se}+P_{ue}^2=0.015+(0.05)^2=0.015+0.0025=0.0175](https://tex.z-dn.net/?f=P_1%3DP_%7Bse%7D%2BP_%7Bue%7D%5E2%3D0.015%2B%280.05%29%5E2%3D0.015%2B0.0025%3D0.0175)
b) The probability that both suppliers will be disrupted using option 2:
![P_2=P_{se}+P_{ue}^2=0.002+(0.13)^2=0.002+0.0169=0.0189](https://tex.z-dn.net/?f=P_2%3DP_%7Bse%7D%2BP_%7Bue%7D%5E2%3D0.002%2B%280.13%29%5E2%3D0.002%2B0.0169%3D0.0189)
Pue = probability of a unique event
Pse = probability of a superevent