M^1/3 divided by m^1/5 = m^2/15<span>
m^1/5 divided by m^1/3 = 1/ </span>m^2/15<span>
m^1/3*m^1/5 = m^8/15
(m^1/3)^1/5 = m^1/15
(m^1/3m^1/5)^0 =1</span>
Answer:
Question one: 3rd option
(Pt. 2 written):
b=ac^2
b=1/8 (4)^2
b= 2
Question two: It is option 2
(Pt. 2 written):
v= m-k
v= -2+7
v= 5
Question three: option 2
(Pt. 2 written):
q=ps/r
q=14(-4)/-7
q=-56/-7
q=8
Question four: It is choice number 3
(Pt. 2 written):
m=3n/4
m=3(8)/4
m=24/4
m=6
Question five: It is choice number 2
(Pt. 2 written):
x=wy+z
x=-8(-5)+6
x=40+6
x=46
Simplify,,,,,,,,,,,,,,,,,,,
tatuchka [14]
It is D , if you plus into a calculator it simplifies for you !!
Answer:
The number of FTE needed to be hired to cover the additional procedures per 7.5 day work is:
= 1.
Step-by-step explanation:
a) Data and Calculations:
Endoscopies EP Studies Total
Projected number of studies 60 5 65
Time in minutes for per study 5 15
Total time required (in minutes) 300 75 375
Full-time Employee (FTE) works 7.5-hour day or 450 minutes (7.5 * 60) per day.
b)The number of full-time employees (FTE) needed to be hired to cover the additional procedures per 7.5-day work is 1. The FTE has 450 minutes available for the additional procedures. She will only require 375 minutes. Therefore, she has extra (idle) time of 75 (450 - 375) minutes per workday.
