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kirza4 [7]
3 years ago
7

A crew will arrive in one week and begin filming a city for a movie. The mayor is desperate to clean the city streets before fil

ming begins. Two teams are​ available, one that requires 200 hours and one that requires "400" hours. If the teams work​ together, how long will it take to clean all of the​ streets? Is this enough time before the cameras begin​ rolling?
Mathematics
1 answer:
Zigmanuir [339]3 years ago
6 0

Answer:

See below in bold.

Step-by-step explanation:

You work in fractions of the city streets done per hour:

1/200 + 1/400 = 1 /x  where x is the number of hours taken by 2 teams.

Multiply through by the LCM 400x:

2x + x = 400

3x = 400

x =  133.33 hours.

As there are 168 hours in a week they will have enough time.

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Deandra completed the first 6 problems at a rate of 3 problems per hour and the last 12 at a rate of 4 problems per hour.

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6 problems ÷ 2 hours = 3 problems per hour

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Determine which ordered pair is a solution of y=8X
Amanda [17]
<span>A.(0,8) B.(-1,8) C.(1.5,10) D.(2,16)

Simply substitute the value of x to get the value of y.

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Suppose you and a friend each choose at random an integer between 1 and 8, inclusive. For example, some possibilities are (3,7),
Bezzdna [24]

Answer and explanation:

Given : Suppose you and a friend each choose at random an integer between 1 and 8, inclusive.

The sample space is

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)  (1,7) (1,8)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)   (2,7) (2,8)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)   (3,7) (3,8)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)   (4,7) (4,8)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)   (5,7) (5,8)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)   (6,7) (6,8)

(7,1) (7,2) (7,3) (7,4) (7,5) (7,6)   (7,7) (7,8)

(8,1) (8,2) (8,3) (8,4) (8,5) (8,6)   (8,7) (8,8)

Total number of outcome = 64

To find : The following probabilities ?

Solution :

The probability is given by,

\text{Probability}=\frac{\text{Favorable outcome }}{\text{Total outcome}}

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The favorable outcome is (5,8)= 1

\text{Probability}=\frac{1}{64}

b) p(sum of the two numbers picked is < 4)

The favorable outcome is (1,1), (1,2), (2,1)= 3

\text{Probability}=\frac{3}{64}

c) p(both numbers match)

The favorable outcome is (1,1), (2,2), (3,3), (4,4), (5,5), (6,6), (7,7), (8,8) = 8

\text{Probability}=\frac{8}{64}

\text{Probability}=\frac{1}{8}

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