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Ksju [112]
3 years ago
13

Helppppppppppppppppppppppppp

Mathematics
1 answer:
Solnce55 [7]3 years ago
8 0
Set the two expressions equal to each other.

7x + 3 = 9x

Simplify, isolate the x, subtract 7x from both sides

7x (-7x) + 3 = 9x (-7x)
3 = 2x

Set the equation = 0, subtract 3 from both sides

3 (-3) = 2x (-3)

0 = 2x - 3

2x - 3, or D, should be your answer

hope this helps
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(x+40)(x+40) show steps
pishuonlain [190]

(x + 40) * (x + 40) =

(x + 40)² =

x² + 80x + 1600

6 0
3 years ago
On Monday, the Dairy Barn sold 14 waffle cones for every 6 sugar cones. On Tuesday, the Dairy Barn sold 7 waffle cones for every
dimulka [17.4K]

Answer:

Option b

Step-by-step explanation:

To represent this proportion using fractions, let the numerators be the number of waffles sold and denominator, the number of sugar cones.

Thus, the proportion that can be used to represent the cone sales is

14/6 = 7/3

3 0
3 years ago
Read 2 more answers
Answer and explanation plz ???
Nutka1998 [239]

Answer:

25

Step-by-step explanation:

Formula

2xy + 1

Givens

x = 3

y = 4

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2*(3)*(4) + 1

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7 0
3 years ago
Help me thanks<br> Xxxxxxxxxx
34kurt

Answer:

A : 6.5

Step-by-step explanation:

7 0
3 years ago
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A factory worker productivity is normally distributed. one worker produces an average of 75 units per day with a standard deviat
Angelina_Jolie [31]
Let Xi be the random variable representing the number of units the first worker produces in day i.
Define X = X1 + X2 + X3 + X4 + X5 as the random variable representing the number of units the
first worker produces during the entire week. It is easy to prove that X is normally distributed with mean µx = 5·75 = 375 and standard deviation σx = 20√5.
Similarly, define random variables Y1, Y2,...,Y5 representing the number of units produces by
the second worker during each of the five days and define Y = Y1 + Y2 + Y3 + Y4 + Y5. Again, Y is normally distributed with mean µy = 5·65 = 325 and standard deviation σy = 25√5. Of course, we assume that X and Y are independent. The problem asks for P(X > Y ) or in other words for P(X −Y > 0). It is a quite surprising fact that the random variable U = X−Y , the difference between X and Y , is also normally distributed with mean µU = µx−µy = 375−325 = 50 and standard deviation σU, where σ2 U = σ2 x+σ2 y = 400·5+625·5 = 1025·5 = 5125. It follows that σU = √5125. A reference to the above fact can be found online at http://mathworld.wolfram.com/NormalDifferenceDistribution.html.
Now everything reduces to finding P(U > 0) P(U > 0) = P(U −50 √5125 > − 50 √5125)≈ P(Z > −0.69843) ≈ 0.757546   .
5 0
3 years ago
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