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kifflom [539]
2 years ago
10

A shopkeeper sold 20 mobile sets of same brand for Rs 3,00.0000 at the same rate in a week What is the selling price of each mob

ile set​
Mathematics
1 answer:
algol [13]2 years ago
7 0

Answer:

15

Step-by-step explanation:

cost of 20 mobile set=300

cost of 1 mobile set

<em>=</em><em>cost </em><em>of </em><em>mobile </em><em>set/</em><em>number </em><em>of </em><em>mobile </em><em>set</em>

<em>=</em><em>3</em><em>0</em><em>0</em><em>/</em><em>2</em><em>0</em>

<em>=</em><em>1</em><em>5</em>

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Please answer correctly! I will mark you as Brainliest!
Hitman42 [59]

Multiply the length by width by the height/3

7 x 6 x 8/3 = 112 in^2

The answer is 112 in^2

8 0
3 years ago
Read 2 more answers
In a two-digit number, the units digit is twice the tens digit. If the number is doubled, it will be 9 more than the number reve
taurus [48]
10(2x)+x 
21x is the original number

12x is the reversed number
24x is the reversed then doubled number

Set up an equation:
24x=21x+9
3x=9
x=3

The original number is 63 and the reversed number is 36, which then is 72 when doubled.
5 0
3 years ago
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How do I work out this problem?
cluponka [151]
First we need to find ∡PRQ. To do this we can use the fact that the sum of the interior angles of a triangle is 180°:

180=R+P+Q
180=R+40+105
180=R+145
R=180-145
R=35

Now, we can use the fact that a strait line (like SRQ) is also 180°:
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4 0
4 years ago
HALP PLEASEEEEEJHEHEHEHE​
laiz [17]

Answer:

Hello! answer B

What I did was concert them to percents so

1/8 is 12.5%

1/4 is 25%

1/2 is 50% so this clearly goes from least to greatest. Hope that helps!

5 0
3 years ago
Astronomers treat the number of stars X in a given volume of space as Poisson random variable. The density of stars in the Milky
cupoosta [38]

Answer:

Explained below.

Step-by-step explanation:

The random variable <em>X</em> is defined as the number of stars in a given volume of space.

X\sim \text{Poisson}\ (\lambda=1)

The probability mass function of <em>X</em> is:

p_{X}(x)=\frac{e^{-\lambda}\lambda^{x}}{x!}

(7)

Compute the probability of exactly two stars in 16 cubic light-years as follows:

P(X=2)=\frac{e^{-1}\times 1^{2}}{2!}=\frac{e^{-1}}{2}=\frac{0.36788}{2}=0.18394\approx 0.184

(8)

Compute the probability of three or more stars in 16 cubic light-years as follows:

P(X\geq 3)=1-P(X

(9)

In 16 cubic light years there is only 1 star.

Then in 1 cubic light years there will be, (1/16) stars.

Then in 4 cubic light years there will be, 4 × (1/16) = (1/4) stars.

(10)

In 16 cubic light years there is only 1 star.

Then in 1 cubic light years there will be, (1/16) stars.

Then in <em>t</em> cubic light years there will be, [<em>t</em> × (1/16)] stars.

7 0
3 years ago
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