Bivariate data contains two variables
Answer:
It is B
Step-by-step explanation:
Answer:
Step-by-step explanation:
28=28
Answer:
Hence the probability that a randomly selected wire from company A's production will meet the specifications is 0.95455.
Step-by-step explanation:
a)![P(0.14 < x < 0.16 ) = P[(0.14 - 0.15)/ 0.005) < (x - \mu) /\sigma < (0.16 - 0.15) / 0.005) ]](https://tex.z-dn.net/?f=P%280.14%20%3C%20x%20%3C%200.16%20%29%20%3D%20P%5B%280.14%20-%200.15%29%2F%200.005%29%20%3C%20%28x%20-%20%5Cmu%29%20%2F%5Csigma%20%20%3C%20%280.16%20-%200.15%29%20%2F%200.005%29%20%5D)
![=P(0.14 < x < 0.16 ) = P[(0.14 - 0.15)/ 0.005)< ((x - 0.15) /0.005) < (0.16 - 0.15) / 0.005) ]](https://tex.z-dn.net/?f=%3DP%280.14%20%3C%20x%20%3C%200.16%20%29%20%3D%20P%5B%280.14%20-%200.15%29%2F%200.005%29%3C%20%28%28x%20-%200.15%29%20%2F0.005%29%20%3C%20%280.16%20-%200.15%29%20%2F%200.005%29%20%5D)
Using z table,
= 0.9773 - 0.02275
= 0.95455.
Answer:
The distance between the two
points on the map
is 225 units
Step-by-step explanation:
Firstly, we need to calculate the distance in units between the two locations on the map
What we have to use here is the distance formula
Mathematically, we have this as;
D = √(x2-x1)^2 + (y2-y1)^2
(x1,y1) = (2,10.5)
(x2,y2) = (14,16)
we have the distance as follows;
D = √(14-2)^2 + (16-10.5)^2
D = √(12^2 + 6.5^2)
D = √(186.25)
D = 13.65 units
now, from the question, 1 unit on the map represents a distance of 16.5 miles
thus, a distance of 13.65 units will be;
16.5 * 13.65 = 225.225
= 225 units
