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IrinaK [193]
4 years ago
13

In 4, days Marsha withdrew $65.80. How much money did Marsha withdraw each day?

Mathematics
1 answer:
Ket [755]4 years ago
5 0

Answer:

$ 16.45

Step-by-step explanation:

Divide 65.80 with 4.

4 days = 65.80

1 day = 16.45

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Which equation is parallel to the graph of y = 2x-5 and goes through the point (3,9)?
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y = 2x + 3

Step-by-step explanation:

the slope is 2 so the line parallel will have a slope of 2

y-9 = 2(x-3)

y-9= 2x - 6

y = 2x +3

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What is the answer to the problem
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The table contains the proof of the theorem of the relationship between slopes of parallel lines. What is the missing statement
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A builder of houses needs to order some supplies that have a waiting time Y for delivery, with a continuous uniform distribution
xenn [34]

Answer:

The Expected cosy of the builder is $433.3

Step-by-step explanation:

$400 is the fixed cost due to delay.

Given Y ~ U(1,4).

Calculating the Variable Cost, V

V = $0 if Y≤ 2

V = 50(Y-2) if Y > 2

This can be summarised to

V = 50 max(0,Y)

Cost = 400 + 50 max(0, Y-2)

Expected Value is then calculated by;

Waiting day =2

Additional day = at least 1

Total = 3

E(max,{0, Y - 2}) = integral of Max {0, y - 2} * ⅓ Lower bound = 1; Upper bound = 4, (4,1)

Reducing the integration to lowest term

E(max,{0, Y - 2}) = integral of (y - 2) * ⅓ dy Lower bound = 2; Upper bound = 4 (4,2)

E(max,{0, Y - 2}) = integral of (y) * ⅓ dy Lower bound = 0; Upper bound = 2 (2,0)

Integrating, we have

y²/6 (2,0)

= (2²-0²)/6

= 4/6 = ⅔

Cost = 400 + 50 max(0, Y-2)

Cost = 400 + 50 * ⅔

Cost = 400 + 33.3

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5 0
3 years ago
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(3x - 2)^5 =(3x - 2)^2
andreyandreev [35.5K]

Answer:

x=3/2,1

Step-by-step explanation:Given

(3x-2)ˆ5-(3x-2)ˆ2=0

(3x-2)ˆ3(3x-2)ˆ2-(3x-2)ˆ2=0

(3x-2)ˆ2{(3x-2)ˆ3-1}=0

(3x-2)ˆ2=0 Or (3x-2)ˆ3-1=0

3x-2=0 Or(3x-2)ˆ3=1

x=3/2 Or 3x-2=1, x=1

4 0
3 years ago
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