Round 396.6 to 396, because 396/9=44, making a whole number. Hope this helps!!
Answer:
They will have to borrow $ 31,211
(23% of the purchase price)
Step-by-step explanation:
Total price : $135,700
Down payment: 20% of total price
Closing costs : 3% of total price
The Colburns will have to borrow 23% of the total price.
0.23*$135,700 = $ 31,211
A three letter word that represents division in a word problem is " cut "
Answer:
![320 = 10 (2)^{t/60}](https://tex.z-dn.net/?f=%20320%20%3D%2010%20%282%29%5E%7Bt%2F60%7D)
If we divide both sides by 10 we got:
![32 = 2^{t/60}](https://tex.z-dn.net/?f=%2032%20%3D%202%5E%7Bt%2F60%7D)
We can apply natural log on both sides and we got:
![ln (32) = \frac{t}{60} ln(2)](https://tex.z-dn.net/?f=%20ln%20%2832%29%20%3D%20%5Cfrac%7Bt%7D%7B60%7D%20ln%282%29%20)
And solving the value of t we got:
![t = 60 \frac{ln(32)}{ln(2)}= 300](https://tex.z-dn.net/?f=%20t%20%3D%2060%20%5Cfrac%7Bln%2832%29%7D%7Bln%282%29%7D%3D%20300)
So then we can conclude that after t = 300 days we will have approximately 320 rabbits
Step-by-step explanation:
For this case we have the following function:
![P(t) = 10 (2)^{t/60}](https://tex.z-dn.net/?f=%20P%28t%29%20%3D%2010%20%282%29%5E%7Bt%2F60%7D)
Where P is the population of rabbis on day t. And for this case we want to find the value of t when P =320 so we can set up the following equation:
![320 = 10 (2)^{t/60}](https://tex.z-dn.net/?f=%20320%20%3D%2010%20%282%29%5E%7Bt%2F60%7D)
If we divide both sides by 10 we got:
![32 = 2^{t/60}](https://tex.z-dn.net/?f=%2032%20%3D%202%5E%7Bt%2F60%7D)
We can apply natural log on both sides and we got:
![ln (32) = \frac{t}{60} ln(2)](https://tex.z-dn.net/?f=%20ln%20%2832%29%20%3D%20%5Cfrac%7Bt%7D%7B60%7D%20ln%282%29%20)
And solving the value of t we got:
![t = 60 \frac{ln(32)}{ln(2)}= 300](https://tex.z-dn.net/?f=%20t%20%3D%2060%20%5Cfrac%7Bln%2832%29%7D%7Bln%282%29%7D%3D%20300)
So then we can conclude that after t = 300 days we will have approximately 320 rabbits
Answer:
a)
b) ![P(4\leq X\leq 8)=0.1954+0.1563+0.1042+0.0595+0.0298=0.5452](https://tex.z-dn.net/?f=P%284%5Cleq%20X%5Cleq%208%29%3D0.1954%2B0.1563%2B0.1042%2B0.0595%2B0.0298%3D0.5452)
c) ![P(X \geq 8) = 1-P(X](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%208%29%20%3D%201-P%28X%3C8%29%20%3D%201-P%28X%5Cleq%207%29%3D1-%5B0.0183%2B0.0733%2B%200.1465%2B0.1954%2B0.1954%2B0.1563%2B%200.1042%2B0.0595%5D%3D0.0511)
d) ![P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559](https://tex.z-dn.net/?f=P%284%5Cleq%20X%20%5Cleq%206%29%3D0.1954%2B0.1563%2B0.1042%3D0.4559)
Step-by-step explanation:
Let X the random variable that represent the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. We know that
The probability mass function for the random variable is given by:
And f(x)=0 for other case.
For this distribution the expected value is the same parameter
,
, ![Sd(X)=2](https://tex.z-dn.net/?f=Sd%28X%29%3D2)
a. Compute both P(X≤4) and P(X<4).
Using the pmf we can find the individual probabilities like this:
![P(X=1)=\frac{e^{-4} 4^1}{1!}=0.0733](https://tex.z-dn.net/?f=P%28X%3D1%29%3D%5Cfrac%7Be%5E%7B-4%7D%204%5E1%7D%7B1%21%7D%3D0.0733)
![P(X=2)=\frac{e^{-4} 4^2}{2!}=0.1465](https://tex.z-dn.net/?f=P%28X%3D2%29%3D%5Cfrac%7Be%5E%7B-4%7D%204%5E2%7D%7B2%21%7D%3D0.1465)
![P(X=3)=\frac{e^{-4} 4^3}{3!}=0.1954](https://tex.z-dn.net/?f=P%28X%3D3%29%3D%5Cfrac%7Be%5E%7B-4%7D%204%5E3%7D%7B3%21%7D%3D0.1954)
![P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954](https://tex.z-dn.net/?f=P%28X%3D4%29%3D%5Cfrac%7Be%5E%7B-4%7D%204%5E4%7D%7B4%21%7D%3D0.1954)
b. Compute P(4≤X≤ 8).
![P(4\leq X\leq 8)=P(X=4)+P(X=5)+ P(X=6)+P(X=7)+P(X=8)](https://tex.z-dn.net/?f=P%284%5Cleq%20X%5Cleq%208%29%3DP%28X%3D4%29%2BP%28X%3D5%29%2B%20P%28X%3D6%29%2BP%28X%3D7%29%2BP%28X%3D8%29)
![P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954](https://tex.z-dn.net/?f=P%28X%3D4%29%3D%5Cfrac%7Be%5E%7B-4%7D%204%5E4%7D%7B4%21%7D%3D0.1954)
![P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563](https://tex.z-dn.net/?f=P%28X%3D5%29%3D%5Cfrac%7Be%5E%7B-4%7D%204%5E5%7D%7B5%21%7D%3D0.1563)
![P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042](https://tex.z-dn.net/?f=P%28X%3D6%29%3D%5Cfrac%7Be%5E%7B-4%7D%204%5E6%7D%7B6%21%7D%3D0.1042)
![P(X=7)=\frac{e^{-4} 4^7}{7!}=0.0595](https://tex.z-dn.net/?f=P%28X%3D7%29%3D%5Cfrac%7Be%5E%7B-4%7D%204%5E7%7D%7B7%21%7D%3D0.0595)
![P(X=8)=\frac{e^{-4} 4^8}{8!}=0.0298](https://tex.z-dn.net/?f=P%28X%3D8%29%3D%5Cfrac%7Be%5E%7B-4%7D%204%5E8%7D%7B8%21%7D%3D0.0298)
![P(4\leq X\leq 8)=0.1954+0.1563+ 0.1042+0.0595+0.0298=0.5452](https://tex.z-dn.net/?f=P%284%5Cleq%20X%5Cleq%208%29%3D0.1954%2B0.1563%2B%200.1042%2B0.0595%2B0.0298%3D0.5452)
c. Compute P(8≤ X).
![P(X \geq 8) = 1-P(X](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%208%29%20%3D%201-P%28X%3C8%29%20%3D%201-P%28X%5Cleq%207%29%3D1-%5BP%28X%3D0%29%2BP%28X%3D1%29%2B%20P%28X%3D2%29%2BP%28X%3D3%29%2BP%28X%3D4%29%2BP%28X%3D5%29%2B%20P%28X%3D6%29%2BP%28X%3D7%29%5D)
![P(X \geq 8) = 1-P(X](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%208%29%20%3D%201-P%28X%3C8%29%20%3D%201-P%28X%5Cleq%207%29%3D1-%5B0.0183%2B0.0733%2B%200.1465%2B0.1954%2B0.1954%2B0.1563%2B%200.1042%2B0.0595%5D%3D0.0511)
d. What is the probability that the number of anomalies exceeds its mean value by no more than one standard deviation?
The mean is 4 and the deviation is 2, so we want this probability
![P(4\leq X \leq 6)=P(X=4)+P(X=5)+P(X=6)](https://tex.z-dn.net/?f=P%284%5Cleq%20X%20%5Cleq%206%29%3DP%28X%3D4%29%2BP%28X%3D5%29%2BP%28X%3D6%29)
![P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954](https://tex.z-dn.net/?f=P%28X%3D4%29%3D%5Cfrac%7Be%5E%7B-4%7D%204%5E4%7D%7B4%21%7D%3D0.1954)
![P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563](https://tex.z-dn.net/?f=P%28X%3D5%29%3D%5Cfrac%7Be%5E%7B-4%7D%204%5E5%7D%7B5%21%7D%3D0.1563)
![P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042](https://tex.z-dn.net/?f=P%28X%3D6%29%3D%5Cfrac%7Be%5E%7B-4%7D%204%5E6%7D%7B6%21%7D%3D0.1042)
![P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559](https://tex.z-dn.net/?f=P%284%5Cleq%20X%20%5Cleq%206%29%3D0.1954%2B0.1563%2B0.1042%3D0.4559)