Answer:
Step-by-step explanation:
Answer:
Rs 120.
Step-by-step explanation:
10=0.85SP-CP; CP+10=0.85SP; SP=[CP+10]/0.85 Eq 1. Let SP= Selling Price and CP= Cost Price
-2 =0.75SP-CP; 0.75SP=C-2; SP=[CP-2]/0.75 Eq 2
[CP+10]/0.85=[CP-2]/0.75 : SP of Eq 1=SP of Eq 2
0.75[CP+10]=0.85[CP-2]
0.75CP+7.5=0.85CP-1.7
0.85CP-0.75CP=-1.7–7.5=9.2
0.10CP=9.2; CP=9.2/0.10
CP=Rs 92 Cost Price of pen
10=0.85SP-92; 0.85SP=92+10=102; SP=102/0.85=Rs 120 Marked Price of pen (answer)
From Eq2: -2=0.75SP-CP; 0.75SP=CP-2=92–2=90; SP=90/0.75=Rs120; -2=0.75(120)-CP; CP-2=0.75(120); CP-2=90; CP=90+2=Rs 92
Set CP of Eq 1=CP of Eq 2:
CP=0.85SP-10 from Eq 1; CP=0.75SP+2 from Eq 2;
0.85SP-10=0.75SP+2; 0.85SP-0.75SP=10+2=12
0.10SP=12; SP=12/0.10=Rs120 is the Marked Price(answer)
Normally, the Selling Price is the marked price. The seller will not disclose the Cost Price because it is the price when the item was acquired or procured, otherwise the buyer will ask for more discounts and based his buying price from the Cost Price if it is known. The calculated SP and CP satisfy both Eq 1 and Eq 2. Both Eq 1 and Eq 2 satisfy the given conditions of the problem above.
When you have two mutually exclusive events, to find the probability of one or another, you add the probabilities.
65+7=72
Final answer: B
Answer:
Step-by-step explanation:
Given data:
SS={0,1,2,3,4}
Let probability of moving to the right be = P
Then probability of moving to the left is =1-P
The transition probability matrix is:
![\left[\begin{array}{ccccc}1&P&0&0&0\\1-P&1&P&0&0\\0&1-P&1&P&0\\0&0&1-P&1&P\\0&0&0&1-P&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%26P%260%260%260%5C%5C1-P%261%26P%260%260%5C%5C0%261-P%261%26P%260%5C%5C0%260%261-P%261%26P%5C%5C0%260%260%261-P%261%5Cend%7Barray%7D%5Cright%5D)
Calculating the limiting probabilities:
π0=π0+Pπ1 eq(1)
π1=(1-P)π0+π1+Pπ2 eq(2)
π2=(1-P)π1+π2+Pπ3 eq(3)
π3=(1-P)π2+π3+Pπ4 eq(4)
π4=(1-P)π3+π4 eq(5)
π0+π1+π2+π3+π4=1
π0-π0-Pπ1=0
→π1 = 0
substituting value of π1 in eq(2)
(1-P)π0+Pπ2=0
from
π2=(1-P)π1+π2+Pπ3
we get
(1-P)π1+Pπ3 = 0
from
π3=(1-P)π2+π3+Pπ4
we get
(1-P)π2+Pπ4 =0
from π4=(1-P)π3+π4
→π3=0
substituting values of π1 and π3 in eq(3)
→π2=0
Now
π0+π1+π2+π3+π4=0
π0+π4=1
π0=0.5
π4=0.5
So limiting probabilities are {0.5,0,0,0,0.5}
Which value is possible for the length of the rectangle?<span>d. 10 in.</span>
