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babymother [125]
3 years ago
11

The Des Plaines River Bridge is 2.1 km long. Engineers are considering re-lighting the bridge with a pole placed every 100 feet

of the bridge length starting from the south end. Each pole will support four light globes. How many globes will be required for the re-lighting project?
Mathematics
2 answers:
Basile [38]3 years ago
8 0
228
hope this helped!
kap26 [50]3 years ago
5 0
Now, there are 3280.84 feet in 1 kilometer, so, how many feet are there in 2.1km? well, 2.1 * 3280.84 or 6889.764 feet.

now, they're putting globes on intervals of 100ft, however, how many times does 100 go into 6889.764? well 6889.764/100  which is rounded up 68.9.

so, that means roughly 69 poles, however, the division excludes the starting point, so if we include that, because is lighting and thus it needs to include the starting point of the bridge, so that'd be 70 poles then.

each pole with 4 globes, 70 * 4 globes then.
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Match each pair of equivalent fractions. 1 . 4/16 2 . 12/18 3 . 4/10 4 . 16/20
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3 0
3 years ago
URGENT please help!!! I need a explanation please!! You own a restaurant and can reopen as long as people seated at a table are
hram777 [196]

Answer:

1) Maximum number of tables that can be kept in the dinning room is 10 tables

2)The maximum number of customers, given 6 people per table is 60 customers

Step-by-step explanation:

The parameters given are;

Distance between people seated at different tables = 6 ft

Dimension of table = 2 m by 1.5 m

Number of people at a table = 6 people

Dimension of dinning;

Width = 12 m

Length = 20 m

Dimension of entrance;

Width = 20/21*30  m

Length = 10 m

With 6 meters between tables and a table with of 1.5, we have for the arrangement in the question;

3 tables = 4.5 m

Distance between = 2 × 6 = 12

4.5 + 12 = 16 m (More room required)

With two tables, we have;

Width of tables = 2×1.5 = 3 m

Distance between = 6 m

Dining room width required = 3 + 6 = 9 m

Therefore, the maximum tables in each row = 2

Given that the dining room area extends to the entrance area, total length = 30 m.

With 4 tables we have;

Length = 4 × 2 + 6 × 3 = 26

While on the entrance side of the dinning room area which is 20 m, we have 3 tables;

Length = 3 × 2 + 6 × 2 = 18

Therefore, both arrangements are acceptable;

Just on entering by the right the length = 20/21×30 m

Therefore, with 3 tables will required number due to proximity wit the door and tables arranged along the right wall of the dinning

1) Maximum number of tables that can be kept in the dinning room = 4 + 3 + 3 = 10 tables

2)The maximum number of customers, given 6 people per table = 6×10 = 60 customers.

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