.75 is the decimal form of 3/4
Answer:
Probability that the sample mean comprehensive strength exceeds 4985 psi is 0.99999.
Step-by-step explanation:
We are given that a random sample of n = 9 structural elements is tested for comprehensive strength. We know the true mean comprehensive strength μ = 5500 psi and the standard deviation is σ = 100 psi.
<u><em>Let </em></u>
<u><em> = sample mean comprehensive strength</em></u>
The z-score probability distribution for sample mean is given by;
Z =
~ N(0,1)
where,
= population mean comprehensive strength = 5500 psi
= standard deviation = 100 psi
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.
Now, Probability that the sample mean comprehensive strength exceeds 4985 psi is given by = P(
> 4985 psi)
P(
> 4985 psi) = P(
>
) = P(Z > -15.45) = P(Z < 15.45)
= <u>0.99999</u>
<em>Since in the z table the highest critical value of x for which a probability area is given is x = 4.40 which is 0.99999, so we assume that our required probability will be equal to 0.99999.</em>
Additive indentity

n = 1
a variable is always a number itself remember 7+n
5+2 = 7
Answer:
x=84
Step-by-step explanation:
Add 4 on both sides
3=x/12-4
7=x/12
Multiply both sides by 12
(12)7=x/12(12)
84=x
Hope this helps! :)
Here we must see in how many different ways we can select 2 students from the 3 clubs, such that the students <em>do not belong to the same club. </em>We will see that there are 110 different ways in which 2 students from different clubs can be selected.
So there are 3 clubs:
- Club A, with 10 students.
- Club B, with 4 students.
- Club C, with 5 students.
The possible combinations of 2 students from different clubs are
- Club A with club B
- Club A with club C
- Club B with club C.
The number of combinations for each of these is given by the product between the number of students in the club, so we get:
- Club A with club B: 10*4 = 40
- Club A with club C: 10*5 = 50
- Club B with club C. 4*5 = 20
For a total of 40 + 50 + 20 = 110 different combinations.
This means that there are 110 different ways in which 2 students from different clubs can be selected.
If you want to learn more about combination and selections, you can read:
brainly.com/question/251701