Answer:
.
Step-by-step explanation:
It is given that a number, x, rounded to 2 significant figures is 1300.
It is possible if,
1. The value of x is greater than of equal to 1250 and less than or equal to 1300.
i.e., 
...(1)
2. The value of x is greater than of equal to 1300 and less than 1350.
i.e., 
...(2)
On combining (1) and (2), we get
1350 is not included in the error interval for x.
Interval notation is .
Therefore, the error interval for x is .
PQ = 9
QR = 28
Note, there is a Q in each line
Combine the two lines: PQ + QR = PR
plug in the numbers to corresponding variables
9 + 28 = PR
PR = 37
hope this helps
Event: Probability: A. Too much enamel 0.18 B. Too little enamel 0.24 C. Uneven application 0.33 D. No defects noted 0.47
let P(AC) = x, P(BC) = y, then P(A) + P(B) + P(C) - (x+y) = 1-0.47 = 0.53 x+y = 0.22
3. The probability of paint defects that results to <span>an improper amount of paint and uneven application? </span>
P(A U B U C) = 0.53
4. <span>the probability of a paint defect that results to</span>
<span>the proper amount of paint, but uneven application?</span>
P(C) - P(AC) - P(BC) = 0.47 - 0.22 = 0.25
A and B are disjoint so P(ABC) = 0, but you can have P(AC) and P(BC). you can't compute these separately here, but you can compute P(AC) + P(BC). By the way, P(AC) eg is just an abbreviated version of P(A∩C).
Try this option (see the attachment), if it is possible check result in other sources.
Lying in the same straight line i think..
