Answer:
Number of calls expected in next week by manager = 7940
Average Number of calls that call center agent will attend in an hour =7 calls
It is also given that, Call center remain open for 10 hours 5 days a week.
Also, it is given that, full time agents work 40 hours a week but are only on call for 35 hours per week ,Part time agents work 20 hours a week but are only on calls 17 hours per week .
⇒Number of hours worked by full time agents × Number of calls attended in an hour × Number of full time agents + Number of hours worked by Part time agents × Number of calls attended in an hour × Number of Part time agents ≤ 7940
⇒35 × 7×Number of full time agents +17 × 7 ×Number of Part time agents ≤ 7940
Option A
⇒35×15×7+17×7×15
= 3675+1785
= 5460
Option B
⇒35 ×7×20+17×7×7
=4900 +833
= 5733
Option C
⇒35×20×7 +17×20×7
=4900+2380
=7280
Option D
⇒25 × 35×7+17×7×5
=6125 +595
=6720
Option E
⇒28×35×7+17×7×10
=6860+1190
=8050
Option E, ⇒ 28 full time agents and 10 part time agents , is best to meet the scheduling needs is most appropriate, that is nearer to 7940 calls.
Answer:
Step-by-step explanation:
Let the length of the rectangle be 'L'
And the width of the rectangle be 'W'
According to the statement:
Perimeter of a rectangle= 2(Length+Width)
Answer:
a = 3.9
Step-by-step explanation:
Isolate the varible by dividing each side by factors that don't contain the variable.
∫(t = 2 to 3) t^3 dt
= (1/4)t^4 {for t = 2 to 3}
= 65/4.
----
∫(t = 2 to 3) t √(t - 2) dt
= ∫(u = 0 to 1) (u + 2) √u du, letting u = t - 2
= ∫(u = 0 to 1) (u^(3/2) + 2u^(1/2)) du
= [(2/5) u^(5/2) + (4/3) u^(3/2)] {for u = 0 to 1}
= 26/15.
----
For the k-entry, use integration by parts with
u = t, dv = sin(πt) dt
du = 1 dt, v = (-1/π) cos(πt).
So, ∫(t = 2 to 3) t sin(πt) dt
= (-1/π) t cos(πt) {for t = 2 to 3} - ∫(t = 2 to 3) (-1/π) cos(πt) dt
= (-1/π) (3 * -1 - 2 * 1) + [(1/π^2) sin(πt) {for t = 2 to 3}]
= 5/π + 0
= 5/π.
Therefore,
∫(t = 2 to 3) <t^3, t√(t - 2), t sin(πt)> dt = <65/4, 26/15, 5/π>.
