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Ostrovityanka [42]
2 years ago
13

- A random drawing will determine which 3 people in a group of 9 will win concert tickets.

Mathematics
1 answer:
Otrada [13]2 years ago
4 0

Answer:

1/3

Step-by-step explanation:

3/9=1/3

the chance of winning concert tickets is 3/9, if you simplify that or do 3/9

(3 divided by 9) it will be 1/3.  

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5 + n2 &gt; 8<br> A. n &gt; 6<br> B. n&gt;3<br> C. n&gt;1.5<br> D. n &lt; 26
san4es73 [151]
N•2>8
then dive both sides 2n>8
and get n>4
4 0
3 years ago
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Solve for x: −2(x + 3) = −2(x + 1) − 4
mash [69]
-2(x + 3) = -2(x + 1) - 4
-2x - 6 = -2x - 2 - 4
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4 0
3 years ago
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Andrea makes 45 tacos for her party. If there are 9 people
jok3333 [9.3K]

Answer:

Each get 5

Step-by-step explanation:

6 0
2 years ago
Suppose the supply function for a certain item is given by S(q)= (q+6)2 and the demand funtion is given by D(q)= (1000)/(q+6).
nasty-shy [4]

Answer: The equilibrium point is where; Quantity supplied = 100 and Quantity demanded = 100

Step-by-step explanation: The equilibrium point on a demand and supply graph is the point at which demand equals supply. Better put, it is the point where the demand curve intersects the supply curve.

The supply function is given as

S(q) = (q + 6)^2

The demand function is given as

D(q) = 1000/(q + 6)

The equilibrium point therefore would be derived as

(q + 6)^2 = 1000/(q + 6)

Cross multiply and you have

(q + 6)^2 x (q + 6) = 1000

(q + 6 )^3 = 1000

Add the cube root sign to both sides of the equation

q + 6 = 10

Subtract 6 from both sides of the equation

q = 4

Therefore when q = 4, supply would be

S(q) = (4 + 6)^2

S(q) = 10^2

S(q) = 100

Also when q = 4, demand would be

D(q) = 1000/(4 + 6)

D(q) = 1000/10

D(q) = 100

Hence at the point of equilibrium the quantity demanded and quantity supplied would be 100 units.

7 0
3 years ago
Find the error with this proof and explain how it mat be corrected in order to clearly prove the equation.
enyata [817]

Answer:

Step-by-step explanation:

It is true that for any given odd integer, square of that integer will also be odd.

i.e if m is and odd integer then m^{2} is also odd.

In the given proof the expansion for (2k + 1)^{2} is incorrect.

By definition we know,

(a+b)^{2} = a^{2} + b^{2} + 2ab

∴ (2k + 1)^{2} = (2k)^{2} + 1^{2} + 2(2k)(1)\\(2k + 1)^{2} = 4k^{2} + 1 + 4k

Now, we know 4k^{2} and 4k will be even values

∴4k^{2} + 1 + 4k will be odd

hence (2k + 1)^{2} will be odd, which means m^{2} will be odd.

7 0
3 years ago
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