What is your question?????
I found this online, hope it helps
Just follow these steps:
Multiply normally, ignoring the decimal points.
Then put the decimal point in the answer - it will have as many decimal places as the two original numbers combined.
Answer:
0.1587
Step-by-step explanation:
Let X be the commuting time for the student. We know that
. Then, the normal probability density function for the random variable X is given by
. We are seeking the probability P(X>35) because the student leaves home at 8:25 A.M., we want to know the probability that the student will arrive at the college campus later than 9 A.M. and between 8:25 A.M. and 9 A.M. there are 35 minutes of difference. So,
= 0.1587
To find this probability you can use either a table from a book or a programming language. We have used the R statistical programming language an the instruction pnorm(35, mean = 30, sd = 5, lower.tail = F)
Answer:
x > -3/7
Step-by-step explanation:
6x - 2(x + 2) > 2 - 3(x + 3)
Distribute the two and three inside the parenthesis.
6x - 2x - 4 > 2 - 3x - 9
Combine like terms.
4x - 4 > -3x - 7
Add 4 to both sides.
4x > -3x - 3
Add 3x to both sides.
7x > -3
Divide both sides by 7.
x > -3/7
Answer:
c) 0.932
99% confidence interval for average weights of all packages sold in small meat trays.
(0.932 ,1.071)
Step-by-step explanation:
Explanation:-
Given random sample of 35 packages in small meat trays produced weight with an average of 1.01 lbs. and standard deviation of 0.18 lbs.
size of the sample 'n' = 35
mean of the sample x⁻= 1.01lbs
standard deviation of the sample 'S' = 0.18lbs
<u>The 99% confidence intervals are given by</u>

The degrees of freedom γ=n-1 =35-1=34
tₐ = 2.0322
99% confidence interval for average weights of all packages sold in small meat trays

( 1.01 - 0.06183 , 1.01+0.06183)
(0.932 ,1.071)
<u>Final answer</u>:-
<u>99% confidence interval for average weights of all packages sold in small meat trays.</u>
<u>(0.932 ,1.071)</u>