Answer:
Switch .91 and .19 and you're good
Step-by-step explanation:
<h3><u>The function that represents the price P of the car after x years is:</u></h3>
![P = 25000(0.935)^x](https://tex.z-dn.net/?f=P%20%3D%2025000%280.935%29%5Ex)
<em><u>Solution:</u></em>
<em><u>The decreasing function is given as:</u></em>
![y = a(1 - r)^t](https://tex.z-dn.net/?f=y%20%3D%20a%281%20-%20r%29%5Et)
Where,
y is future value
a is initial value
r is decreasing rate in decimal
t is time period
From given,
a = 25000
![r = 6.5 \% = \frac{6.5}{100} = 0.065](https://tex.z-dn.net/?f=r%20%3D%206.5%20%5C%25%20%3D%20%5Cfrac%7B6.5%7D%7B100%7D%20%3D%200.065)
number of years = x
future value = P
Therefore,
![P = 25000(1 - 0.065)^x\\\\P = 25000(0.935)^x](https://tex.z-dn.net/?f=P%20%3D%2025000%281%20-%200.065%29%5Ex%5C%5C%5C%5CP%20%3D%2025000%280.935%29%5Ex)
Thus the function is found
Answer:
6x^2^x^2+x^4
Step-by-step explanation:
Answer: ![\bold{\dfrac{7}{8}=87.5\%}](https://tex.z-dn.net/?f=%5Cbold%7B%5Cdfrac%7B7%7D%7B8%7D%3D87.5%5C%25%7D)
<u>Step-by-step explanation:</u>
"At least one girl" means P(1 girl) + P(2 girls) + P(3 girls) or 1 - P(all boys)
I will use the latter: 1 - P(all boys)
P(all boys) = ![\dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}=\dfrac{1}{8}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cdfrac%7B1%7D%7B2%7D%3D%5Cdfrac%7B1%7D%7B8%7D)
![1 - P(all\ boys) \quad = \quad 1-\dfrac{1}{8}\quad =\large\boxed{\dfrac{7}{8}}](https://tex.z-dn.net/?f=1%20-%20P%28all%5C%20boys%29%20%5Cquad%20%3D%20%5Cquad%201-%5Cdfrac%7B1%7D%7B8%7D%5Cquad%20%3D%5Clarge%5Cboxed%7B%5Cdfrac%7B7%7D%7B8%7D%7D)
Answer:
I am not able to answer your question because I am unsure of what to solve for. :/
Step-by-step explanation: