Answer:
<u>hu</u>
<u />
Step-by-step explanation:
Answer:
The number of boxes of lettuce the supermarket should purchase = 273 boxes
Step-by-step explanation:
This is a case of understocking and overstocking.
We are given;
Cost price = $4 per box
Selling price = $10 per box
Salvage value = $2
The formula for cost of understocking is;
C_u = S.P - C.P
Where;
C_u is cost of understocking
S.P is selling price
C.P is cost price
Thus;
C_u = $10 - $4
C_u = $6
Cost of overstocking is given by;
C_o = Cost Price - Salvage value
C_o = $4 - $2
C_o = $2
Now,
P(D ≤ Q) = C_u/(C_u + C_o)
P(D ≤ Q) = 6/(6 + 2) = 6/8 = 0.75
From the z-table i attached, the z value of 0.75 is approximately 0.67
Thus;
The number of boxes of lettuce the supermarket should purchase =
D + (z-value x standard deviation)
We are given;
D = 250
Standard deviation = 34
Thus;
The number of boxes of lettuce the supermarket should purchase = 250 + (0.67 * 34) = 272.78 ≈ 273 boxes
Answer:
Step-by-step explanation:
By exterior angle theorem:
4x° = 2x° + 70°
4x° - 2x° = 70°
2 x° = 70°
2x = 70
x = 70/2
x = 35
Answer:
-2
Step-by-step explanation:
Answer: 0.878
Step-by-step explanation:
Given : P( Internet access) = 80.5%= 0.805
P(cable television) = 82% = 0.82
P( Internet access and cable television) = 74.7% =0.747
Formula: P(A or B)= P(A ) + P(B) - P(A and B)
So, P( Internet access or cable television) = P( Internet access) + P(cable television) -P( Internet access and cable television)
= 0.805+0.82-0.747
= 0.878
The probability that a randomly selected household in the survey had either Internet access or cable = 0.878