Answer:
Step-by-step explanation:
Our inequality is |125-u| ≤ 30. Let's separate this into two. Assuming that (125-u) is positive, we have 125-u ≤ 30, and if we assume that it's negative, we'd have -(125-u)≤30, or u-125≤30.
Therefore, we now have two inequalities to solve for:
125-u ≤ 30
u-125≤30
For the first one, we can subtract 125 and add u to both sides, resulting in
0 ≤ u-95, or 95≤u. Therefore, that is our first inequality.
The second one can be figured out by adding 125 to both sides, so u ≤ 155.
Remember that we took these two inequalities from an absolute value -- as a result, they BOTH must be true in order for the original inequality to be true. Therefore,
u ≥ 95
and
u ≤ 155
combine to be
95 ≤ u ≤ 155, or the 4th option
Cant really see it................
Answer:
Moving the decimal 3 places to the right in a decimal number is the same as multiplying the number by 1000.
A-4=B
2+C=B
A+B+C=13
A=b+4
C=B-2
Plug in to equation 3
(B+4)+B+(B-2)=13
Rearrange
B=11/3
A=(11/3)+4= 23/3
C=(11/4)-2= 3/4
Answer: 0.8461
Step-by-step explanation:
Given : 
Let x be the random variable that represents the cost for the hospital emergency room visit.
We assume that cost for the hospital emergency room visit is normally distributed .
z-score for x=1000 ,
![z=\dfrac{1000-1400}{392}\approx-1.02\ \ \ [\because z=\dfrac{x-\mu}{\sigma}]](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B1000-1400%7D%7B392%7D%5Capprox-1.02%5C%20%5C%20%5C%20%5B%5Cbecause%20z%3D%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5D)
Using z-value table , we have
P-value =P(x>1000)=P(z>-1.02)=1-P(z≤ -1.02)=1-0.1538642
=0.8461358≈0.8461 [Rounded nearest 4 decimal places]
Hence, the probability that the cost will be more than $1000 = 0.8461
