Answer:
inverse relation of x is f⁻(X) = (x+15)/-6
Step-by-step explanation:
Answer:
1. 136.3/53% = 257.16981...
2. 99/231.2 = 0.4282
3. 853.2/51% = 1672.94117....
4. 439.7/0.56 = 785.17857
5. 496.1/0.92 = 539.23913
6. 0.32 x 238.3 = 76.256
7. 99/376.3 = 0.26308
8. 0.08 x 493.7 = 39.496
Step-by-step explanation:
Good luck
Answer:
123
Step-by-step explanation:
if the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadth
Answer:
Two diameters that separate the top 4% and the bottom 4% are 5.77 and 5.53 respectively.
Step-by-step explanation:
We are given that the diameters of bolts produced in a machine shop are normally distributed with a mean of 5.65 millimeters and a standard deviation of 0.07 millimeters.
<em>Let X = diameters of bolts produced in a machine shop</em>
So, X ~ N(
)
The z score probability distribution is given by;
Z = 
~ N(0,1)
where, 
= population mean
= standard deviation
<u>Now, we have to find the two diameters that separate the top 4% and the bottom 4%.</u>
- Firstly, Probability that the diameter separate the top 4% is given by;
P(X > x) = 0.04
P( > 
) = 0.04
P(Z > ) = 0.04
<em>So, the critical value of x in z table which separate the top 4% is given as 1.7507, which means;</em>
= 1.7507
= 5.65 + 0.122549 = 5.77
- Secondly, Probability that the diameter separate the bottom 4% is given by;
P(X < x) = 0.04
P( < ) = 0.04
P(Z < ) = 0.04
<em>So, the critical value of x in z table which separate the bottom 4% is given as -1.7507, which means;</em>
= -1.7507
= 5.65 - 0.122549 = 5.53
Therefore, the two diameters that separate the top 4% and the bottom 4% are 5.77 and 5.53 respectively.
