Consider the continuous random variable x, which has a uniform distribution over the interval from 110 to 150. The probability that x will take on a value between 120 and 125 is .

The probability that x will take on a value between 120 and 125 is 0.14145

Step-by-step explanation:

For uniform distribution between a & b

Mean, xbar = (a + b)/2

Standard deviation, σ = √((b-a)²/12)

For 110 and 150,

Mean, xbar = (150 + 110)/2 = 130

Standard deviation, σ = √((150-110)²/12 = 11.55

To find the probability that x will take on a value between 120 and 125

We need to standardize 120 & 125

z = (x - xbar)/σ = (120 - 130)/11.55 = - 0.87

z = (x - xbar)/σ = (125 - 130)/11.55 = - 0.43

P(120 < x < 125) = P(-0.87 < x < -0.43)

We'll use data from the normal probability table for these probabilities

P(120 < x < 125) = P(-0.87 < x < -0.43) = P(z ≤ -0.43) - P(z ≤ -0.86) = 0.33360 - 0.19215 = 0.14145

Hope this Helps!!!

mark other person brainliest

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204