Assume that x has a normal distribution with the specified mean and standard deviation. find the indicated probability. (round your answer to four decimal places.) μ = 15.7; σ = 3.6 p(10 ≤ x ≤ 26) =

P(10 ≤ x ≤ 26) = 0.94083  
First, let's determine how many standard deviations each of the limits are  
(10 - 15.7) / 3.6 = -1.583333333 
(26 - 15.7) / 3.6 = 2.861111111  
So the probability of x being between 10 and 26 is the area covered by a normal distribution curve from -1.58333 to 2.86111. Using a standard normal table, look up the values -1.58 and 2.86. 
For the curve from infinity on the left to 2.86, you get the value 0.99788 which means that 99.788% of the curve is to the left of that point. 
For the curve from infinity on the left to -1.58, you get the value 0.05705 which means that 5.705% of the curve is to the left of that point. 
Since we want the area between those two points, we can simply subtract.
So: 
0.99788 - 0.05705 = 0.94083 
 So p(10 ≤ x ≤ 26) = 0.94083

Maybe the answer is 20mm. all i really did was just multiply, which is obvious.

1. isolate the variable, in this case, d. subtract 35.5 on both sides of the equation.2. 52 subtracted by 35.5 is 16.53. therefore your answer would be d= 16.5this is the basics of pre algebra, isolating variable through adding, subtracting, multiplying, and dividing, if you have any other questions to ask just comment!