Answer:
6
Step-by-step explanation:
Find the percentage. 5 percent of 10 is 0.5. 50 cents.
25 X 2(AMOUNT OF 5 PERCENT OF A DOLLAR IN A DOLLAR) = 50.
50 hours
A rational number between the two given ones is -0.455, such that:
-0.45 > -0.455 > -0.46
<h3>How to find a rational number between the two given ones?</h3>
A rational number is any number that can be written as a quotient between two integer numbers.
Particularly, any number with a finite number of digits after the decimal point is also a rational number.
So to find a rational number between -0.45 and -0.46 we could se:
-0.455, such that:
-0.45 > -0.455 > -0.46
Learn more about rational numbers:
brainly.com/question/12088221
#SPJ1
Answer:
4
Step-by-step explanation:
Hope this helps!! Have a good day!
Answer:
a.![P(E_1/A)=0.0789](https://tex.z-dn.net/?f=P%28E_1%2FA%29%3D0.0789)
b.
\
c.![P(E_3/A)=0.526](https://tex.z-dn.net/?f=P%28E_3%2FA%29%3D0.526)
Step-by-step explanation:
Let
are the events that denotes the good drive, medium drive and poor risk driver.
![P(E_1)=0.30,P(E_2)=0.50,P(E_3)=0.20](https://tex.z-dn.net/?f=P%28E_1%29%3D0.30%2CP%28E_2%29%3D0.50%2CP%28E_3%29%3D0.20)
Let A be the event that denotes an accident.
![P(A/E_1)=0.01](https://tex.z-dn.net/?f=P%28A%2FE_1%29%3D0.01)
![P(A/E_2=0.03](https://tex.z-dn.net/?f=P%28A%2FE_2%3D0.03)
![P(A/E_3)=0.10](https://tex.z-dn.net/?f=P%28A%2FE_3%29%3D0.10)
The company sells Mr. Brophyan insurance policy and he has an accident.
a.We have to find the probability Mr.Brophy is a good driver
Bayes theorem,![P(E_i/A)=\frac{P(A/E_i)\cdot P(E_1)}{\sum_{i=1}^{i=n}P(A/E_i)\cdot P(E_i)}](https://tex.z-dn.net/?f=P%28E_i%2FA%29%3D%5Cfrac%7BP%28A%2FE_i%29%5Ccdot%20P%28E_1%29%7D%7B%5Csum_%7Bi%3D1%7D%5E%7Bi%3Dn%7DP%28A%2FE_i%29%5Ccdot%20P%28E_i%29%7D)
We have to find ![P(E_1/A)](https://tex.z-dn.net/?f=P%28E_1%2FA%29)
Using the Bayes theorem
![P(E_1/A)=\frac{P(A/E_1)\cdot P(E_1)}{P(E_1)\cdot P(A/E_1)+P(E_2)P(A/E_2)+P(E_3)P(A/E_3)}](https://tex.z-dn.net/?f=P%28E_1%2FA%29%3D%5Cfrac%7BP%28A%2FE_1%29%5Ccdot%20P%28E_1%29%7D%7BP%28E_1%29%5Ccdot%20P%28A%2FE_1%29%2BP%28E_2%29P%28A%2FE_2%29%2BP%28E_3%29P%28A%2FE_3%29%7D)
Substitute the values then we get
![P(E_1/A)=\frac{0.30\times 0.01}{0.01\times 0.30+0.50\times 0.03+0.20\times 0.10}](https://tex.z-dn.net/?f=P%28E_1%2FA%29%3D%5Cfrac%7B0.30%5Ctimes%200.01%7D%7B0.01%5Ctimes%200.30%2B0.50%5Ctimes%200.03%2B0.20%5Ctimes%200.10%7D)
![P(E_1/A)=0.0789](https://tex.z-dn.net/?f=P%28E_1%2FA%29%3D0.0789)
b.We have to find the probability Mr.Brophy is a medium driver
![P(E_2/A)=\frac{0.03\times 0.50}{0.038}=0.395](https://tex.z-dn.net/?f=P%28E_2%2FA%29%3D%5Cfrac%7B0.03%5Ctimes%200.50%7D%7B0.038%7D%3D0.395)
c.We have to find the probability Mr.Brophy is a poor driver
![P(E_3/A)=\frac{0.20\times 0.10}{0.038}=0.526](https://tex.z-dn.net/?f=P%28E_3%2FA%29%3D%5Cfrac%7B0.20%5Ctimes%200.10%7D%7B0.038%7D%3D0.526)