he Greensboro Convention & Visitors Bureau has determined that across all hotels in the Triad region, the price for a one-night stay follows a normal distribution with mean $143 and standard deviation $38. What is the probability that a randomly chosen hotel charges between $120 and $160 per night?

0.4002

Step-by-step explanation:

Given that :

Mean (m) = 143

Standard deviation (σ) = 38

What is the probability that a randomly chosen hotel charges between $120 and $160 per night?

(120 < x < 160)

Obtain the standardized score :

(x - m) / σ

x : 120

Zscore = (120 - 143) / 38 = −0.605263

p(Z < −0.605263) = 0.2725

x : 160

Zscore = (160 - 143) / 38 = 0.4473684

p(Z < 0.4473684) = 0.6727

p(Z < 0.4473684) - p(Z < −0.605263)

= 0.6727 - 0.2725

= 0.4002

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