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3 years ago

Please help me with this question

Mathematics

1 answer:

Strike441 [17]3 years ago

5 0

does the car and bike angles are given?

to find the no of pupils

angles / 360° x 36

walk: 90/360 x 36 = 9 pupils

bus: 129/360 x 36 = 12 pupils

*otherswise if bike and cars have the samw angle therefore 7.5 puipls for bike and cars but i dont think it is possible im so sorry

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8x+112x+392 factor perfect squares

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120x+392

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3 years ago

Simplify: 0.9(2b-1)-0.5b+1

Nutka1998 [239]

Answer:

Step-by-step explanation:

0.9*2b = 1.8b

0.9*-1 = -0.9

So far we have 1.8b-0.9. It can't be simplified further.

Then, we add the 2nd part, -0.5b+1.

We have:

1.8b-0.9-0.5b+1. Next we combine like terms.

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2 years ago

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(2x + 3)3

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6 0

3 years ago

Each flight uses a jet with a capacity of 180 seats. The airline measures the utilization of the aircraft by passenger load. Pas

garri49 [273]

<h2>Answer:</h2>

(a). 5400

(b). 57 - 0.95, 85 - 0.48 and 94 - 0.44

(c). Less load in Flight number 85 and 94.

<h2>Step-by-step explanation:</h2>

In the question,

Number of seats in a jet = 180

And,

$Passenger\,Load=\frac{Number\,of\,seats\,sold}{Number\,of\,available\,seats}$

Now,

Flight Number 57 has seats sold = 5130

Flight Number 85 has seats sold = 2592

Flight Number 94 has seats sold = 2376

(a).

Available seat capacity for each flight number in June can be found out by,

Number of days in the month of June = 30 days

Each day the flight has = 180 seats.

So,

Total number of available seat capacity for each flight in June is = 180 x 30 = 5400

Therefore, available seat capacity for each flight = 5400

(b).

Passenger load for each flight can be given by,

For Flight number 57,

$Passenger\,Load=\frac{Number\,of\,seats\,sold}{Number\,of\,available\,seats}\\Passenger\,Load=\frac{5130}{5400}\\Passenger\,Load=0.95$

For Flight number 85,

$Passenger\,Load=\frac{Number\,of\,seats\,sold}{Number\,of\,available\,seats}\\Passenger\,Load=\frac{2592}{5400}\\Passenger\,Load=0.48$

For Flight number 94,

$Passenger\,Load=\frac{Number\,of\,seats\,sold}{Number\,of\,available\,seats}\\Passenger\,Load=\frac{2376}{5400}\\Passenger\,Load=0.44$

(c).

The recommendation for the Eastern skies can be that the Passenger Load in the Flight number 57 is near to perfect.

But, the Passenger Load in Flight number 85 and 94 is less than required so there is need of selling more seats in the respective airline flights.

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3 years ago