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denpristay [2]
3 years ago
5

An Airbus has 1 first class section, 1 business-class section and 4 economy-class sections. The first class section has 14 seats

. 5/7 are being used. The business-class section has 64 seats. 5/16 are being used. Each economy-class section has 108. 5/9 in each section are being usd. Are more than half seats on the plane being used? Yes Or No Show your working.
HELP!!!!!!!
Mathematics
1 answer:
serious [3.7K]3 years ago
6 0

Answer:

first class section has  14 Seats

Seat used = (5/7) * 14 = 10 Seats

Business class Section has 64 Seats

Seats used = (5/16) * 64 = 20 Seats

Economy class Section has 4 * 108 = 432 Seats

Seats used = (5/9) * 432 = 240 Seats

Total Seats = 14 + 64 + 432 = 510

Total Seats Used = 10 + 20 + 240 = 270

270/510  = 270 ( 2 * 255)   >  1/2

Yes , more than half of the seats being used

Step-by-step explanation:

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The point P(–4, 4) that is \frac{2}{5} of the way from A to B on the directed line segment AB.

Solution:

The points of the line segment are A(–8, –2) and B(6, 19).

P is the point that bisect the line segment in \frac{2}{5}.

So, m = 2 and n = 5.

x_1=-8, y_1=-2, x_2=6, y_2=19

By section formula:

$P(x, y)=\left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right)

$P(x, y)=\left(\frac{2\times 6+5\times (-8)}{2+5}, \frac{2\times 19+5\times (-2)}{2+5}\right)

$P(x, y)=\left(\frac{12-40}{7}, \frac{38-10}{7}\right)

$P(x, y)=\left(\frac{-28}{7}, \frac{28}{7}\right)

P(x, y) = (–4, 4)

Hence the point P(–4, 4) that is \frac{2}{5} of the way from A to B on the directed line segment AB.

4 0
3 years ago
A realtor earns a 3.25% commission
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Since the commission is 3.25%, we simply multiply this fraction in decimal form by x, since it will equal 10566.40:

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6 0
3 years ago
Help with this task it's confusing can u explain how you got the answer
Andru [333]
Oh my goodness !  You were doing such an absolutely beautiful job,
as far as you went, but then you ran into some rough road and quit.

You've got the correct expressions for the ages of the three people:

-- Will . . . w
-- Ben . . . w+3
-- Jan . . . 2(w+3)
You slipped up when you expanded Jan's age:  2(w+3) = <u>2w + 6</u> ,
and it was all down hill from there.

Let's do it again, together:

-- Will . . . w
-- Ben . . . w + 3
-- Jan . . . 2w + 6

Total:  (w + w + 2w) + (3 + 6) = 4w + 9

So the equation is:    <em><u>4w + 9 = 41</u></em>
Now you're supposed to solve it.

Subtract 9 from each side:   4w  =  32

Divide each side by 4:        <u>w = 8</u>

-- Will = w . . . . . 8 y.o.
-- Ben = w+3 . . . 11 y.o.
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When will Jan be twice as old as Will ?
That'll happen in 'x' years.
At that time, Will will be (8+x) and Jan will be (22+x),
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22 + x  =  2(8 + x)

22 + x  =  16 + 2x

Subtract 'x' from each side:    22  =  16 + x

Subtract  16  from each side:  <em>  6 = x</em>

<u>Check:</u>

In 6 years, Jan will be (22+6) = 28,
and Will will be (8+6) = 14 .

28 = twice as old as 14.      yay!

Can I make a little suggestion ?
I'm going to make it anyway:

Your problem was neatness.
You were doing great work in that big open space on the sheet, but it
started to get ragged.  When you tried to look back to see if you made
a mistake, you couldn't find it in the mess.
This is not an easy problem, but you definitely know your stuff. 
I think if you keep it a little neater, you're going to sparkle !
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