Answer: 1/2, 1/2
Step-by-step explanation:
A die has 6 faces.
Total number of faces = 6
Possible outcomes are :1,2,3,4,5,6
Even numbers : 2, 4, and 6 = 3
Non even numbers: 1,3,and 5= 3
First statement:
P(Even number) = 3/ 6 = 1/2
P(X=1) = 1/2
P(odd number) = 3/6 = 1/2
P(X=0) = 1/2
Second statement:
P(2) = 1/6
P(3) = 1/6
P(4) = 1/6
P(2 or 3 or 4) = (1/6) + (1/6) + (1/6)
P(2 or 3 or 4) = (1+1+1)/6 = 3/6 =1/2
Thus,
P(Y=1) = 1/2
Therefore,
P(X =0,Y=1) = 1/2, 1/2
Answer:
2
Step-by-step explanation:
Simplify the following:
(-6 - 4)/(-5)
Hint: | Group the negative terms in -6 - 4 together and factor out the minus sign.
-6 - 4 = -(6 + 4):
(-(6 + 4))/(-5)
Hint: | Evaluate 6 + 4.
6 + 4 = 10:
(-10)/(-5)
Hint: | In (-10)/(-5), the numbers 10 in the numerator and -5 in the denominator have gcd greater than one.
The gcd of 10 and -5 is 5, so (-10)/(-5) = (-(5×2))/(5 (-1)) = 5/5×(-2)/(-1) = (-2)/(-1):
(-2)/(-1)
Hint: | Cancel common terms in the numerator and denominator of (-2)/(-1).
(-2)/(-1) = (-1)/(-1)×2 = 2:
Answer: 2
Answer: 0.0475
Step-by-step explanation:
Let x = random variable that represents the number of a particular type of bacteria in samples of 1 milliliter (ml) of drinking water, such that X is normally distributed.
Given: 
The probability that a given 1-ml will contain more than 100 bacteria will be:
![P(X>100)=P(\dfrac{X-\mu}{\sigma}>\dfrac{100-85}{9})\\\\=P(Z>1.67)\ \ \ \ [Z=\dfrac{X-\mu}{\sigma}]\\\\=1-P(Zz)=1-P(Z](https://tex.z-dn.net/?f=P%28X%3E100%29%3DP%28%5Cdfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%3E%5Cdfrac%7B100-85%7D%7B9%7D%29%5C%5C%5C%5C%3DP%28Z%3E1.67%29%5C%20%5C%20%5C%20%5C%20%5BZ%3D%5Cdfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%5D%5C%5C%5C%5C%3D1-P%28Z%3C1.67%29%5C%20%5C%20%5C%20%5BP%28Z%3Ez%29%3D1-P%28Z%3Cz%29%5D%5C%5C%5C%5C%3D1-%200.9525%3D0.0475)
∴The probability that a given 1-ml will contain more than 100 bacteria
0.0475.
Your question isn’t clear
