The volume of a rectangular prism is given by length x width x height.
Now in the question the volume is given, length is given, and width is given.
Answer:
Probability that event will have both red marbles will be 2.3% or 0.023.
Step-by-step explanation:
Given:
Total red marbles=5
Total blue marbles=25
Total number of marbles =30
To Find:
The probability that both are red without replacement of marbles.
Solution:
Now
Total sample space is 30 and total red marbles are 5
For a event that getting red marble probability is ,
=Total red marbles /total marbles.
=5/30
=1/6
So probability if getting red marble is 1/6
<em>Now for second chance there will be 4 red marbles remaining and 29 total marbles so,</em>
In second chance probability if getting red marble will be
=total red marbles present/total marbles remaining
=4/29
Now ,
The required probability will be getting both at a time
i.e probability getting red AND red marble so here AND operator which means multiple both the probability.
Probability both will have red =1/6*4/29
=4/(29*6)=2/(29*3)
=0.022988
=0.023
=2.3 %
Mean = 0.497 in, SD = 0.003 in
Required diameter ranges between 0.496 in and 0.504 in
Anything other diameter obtained is not acceptable.
That is;
P(x<0.496) and P(x>0.504) are not acceptable.
Now,
P(x<0.496) = P(Z< (0.496-0.497)/0.003)) = P(Z<-0.33)
From Z tables, P(Z<-0.33) = 0.3707
Similarly,
P(x>0.504) = P(Z> (0.504-0.497)/0.003)) = P(Z>2.33)
From Z tables, P(Z>2.33) = 1-0.9901 = 0.0099
Therefore, unacceptable proportion = P(x<0.496)+P(x>0.504) = 0.3707+0.0099 = 0.3806 or 38.06%
Answer:
The steps are numbered below
Step-by-step explanation:
To solve a maximum/minimum problem, the steps are as follows.
1. Make a drawing.
2. Assign variables to quantities that change.
3. Identify and write down a formula for the quantity that is being optimized.
4. Identify the endpoints, that is, the domain of the function being optimized.
5. Identify the constraint equation.
6. Use the constraint equation to write a new formula for the quantity being optimized that is a function of one variable.
7. Find the derivative and then the critical points of the function being optimized.
8. Evaluate the y-values of the critical points and endpoints by plugging them into the function being optimized. The largest y- value is the global maximum, and the smallest y-value is the global minimum.
Answer:
2.04
Step-by-step explanation:
