The equation to determine the weekly earnings of an employee at the hamburger shack is given by w(x), where x is the number of h
ours worked. w(x) = 10x, 0 ≤ x ≤ 40 15(x − 40) + 400, x > 40 determine the difference in salary, in dollars, for an employee who works 52 hours versus one who works 38 hours. determine the number of hours an employee must work in order to earn $445. explain how you arrived at this answer.
1 answer:
To arrive at the answer, I graphed the piecewise function using a graphing calculator, then evaluated it at x=38 and x=52. I drew a line at y=445 and found the point of intersection on the graph.
a) The difference between earnings for 52 hours ($580) and earnings for 38 hours ($380) is $200.
b) An employee must work 43 hours to earn $445.
a) The difference between earnings for 52 hours ($580) and earnings for 38 hours ($380) is $200.
b) An employee must work 43 hours to earn $445.
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Answer:
Option E) 61.6
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 100 bushels per acre
Standard Deviation, σ = 30 bushels per acre
We assume that the distribution of yield is a bell shaped distribution that is a normal distribution.
Formula:
P(X>x) = 0.90
We have to find the value of x such that the probability is 0.90
P(X > x)
Calculation the value from standard normal table, we have,
Hence, the yield of 61.6 bushels per acre or more would save the seed.
Answer:
a)
b)d(t)=500t
c)
Step-by-step explanation:
d = Horizontal distance
s = the distance between the plane and the radar station
The horizontal distance (d), the one mile altitude, and s form a right triangle.
So, use Pythagoras theorem
a)
(b) Express d as a function of the time t (in hours) that the plane has flown.
d(t)=500t
(c) Use composition to express s as a function of t.
using b
First find the tangent line
dy/dx=2x
at x=6, the slope is 2(6)=12
so
use point slope form
y-y1=m(x-x1)
point is (6,36)
so
y-36=12(x-6)
y-36=12x-72
y=12x-36
alright, so we know they intersect at x=6
and y=12x-36 is below y=x^2
so we do
![[\frac{x^3}{3}-6x^2+36x]\limits^6_0=(\frac{6^3}{3}-6(6)^2+36(6))-(0)=\frac{216}{3}-216+216=](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bx%5E3%7D%7B3%7D-6x%5E2%2B36x%5D%5Climits%5E6_0%3D%28%5Cfrac%7B6%5E3%7D%7B3%7D-6%286%29%5E2%2B36%286%29%29-%280%29%3D%5Cfrac%7B216%7D%7B3%7D-216%2B216%3D)
the area under the curve bounded by the lines and the x axis is 72 square units
dy/dx=2x
at x=6, the slope is 2(6)=12
so
use point slope form
y-y1=m(x-x1)
point is (6,36)
so
y-36=12(x-6)
y-36=12x-72
y=12x-36
alright, so we know they intersect at x=6
and y=12x-36 is below y=x^2
so we do
the area under the curve bounded by the lines and the x axis is 72 square units