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avanturin [10]
3 years ago
14

7. Sorav bought a jacket for Rs. 2500 sold it for Rs. 1900. Find her loss.​

Mathematics
2 answers:
Morgarella [4.7K]3 years ago
4 0

Answer:

600

Step-by-step explanation:

if SP>CP,  

It is profit [SP - CP = Profit]

if CP>SP,  

It is loss  [CP - SP = Loss]

 

Here in the question CP>SP so,  

loss = CP - SP  

      =2500 - 1900  

      = 600

Nataliya [291]3 years ago
3 0

Answer:

24

Step-by-step explanation:

C.P=2500

S.P=1900

Here,

Cp>Sp

So,

Loss= cp-sp

---------- ×100%

cp

=2500-1900

----------------- ×100%

2500

= 600

------- ×100 %

2500

= 24.

<h2>If it wrong, then sorry .</h2>

<h2>Hope it helps you.</h2>
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In 1985 there were 275 cell phone subscribers in the small town of China grove. Subscribers increased by 40% each year after 198
emmasim [6.3K]
1985year ->275cell phones->100%
1994year -> x cell phones ->100%+360%
40% • (1994-1985)= 40% • 9= 360%

275 cell phone 100%
X cell phones. 100+360%=460%
x=275•460%/100%
x=12420/100
x=1265
99% sure this is a correct answer


x=275 • 460%/100%
x=12,420/100
x=1265

1994-1985= 9 yrs • 40%= 360%

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Is the graph shown below that of a function?
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Answer:

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Step-by-step explanation:

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3 years ago
Nevia spent $0.75 on 1.5 pounds of bananas. At this rate, how much will 6 pounds of bananas cost?
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2 years ago
Determine n between 0 and 19 such that (2311)(3912) ≡ n mod 20.
sleet_krkn [62]

You can write 2311 and 3912 in the form 20q+r:

2311=115\cdot20+11

3912=125\cdot20+12

Then

2311\cdot3912=(115\cdot20+11)(125\cdot20+12)

2311\cdot3912=115\cdot125\cdot20^2+(11\cdot125+12\cdot115)\cdot20+11\cdot12

Taken modulo 20, the terms containing powers of 20 vanish and you're left with

2311\cdot3912\equiv11\cdot12\equiv132\pmod{20}

We further have

132=6\cdot20+12

so we end up with

2311\cdot3912\equiv12\pmod{20}

and so n=12.

###

If instead you're trying to find 2311^{3912}\pmod{20}, you can apply Euler's theorem. We can show that \mathrm{gcd}(2311,20)=1 using the Euclidean algorithm. Then since \varphi(20)=8, and 8 divides 3912, we have

2311^{3912}\equiv2311^{489\cdot8}\equiv(2311^{489})^8\equiv1\pmod{20}

To show 2311 and 20 are coprime:

2311 = 115*20 + 11

20 = 1*11 + 9

11 = 1*9 + 2

9 = 4*2 + 1   =>  gcd(2311, 20) = 1

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3 years ago
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Answer: that answer is A, assign a number to each family, use a computer program to select 20 random numbers

Step-by-step explanation:

4 0
3 years ago
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