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stiv31 [10]
3 years ago
10

a school bus traveled at an average speed of 35 miles per hour for 45 minutes . what distance did the school bus travel

Mathematics
2 answers:
rodikova [14]3 years ago
5 0
45/60=0.75. Speed*time=distance. 35(0.75)= 26.25 miles
expeople1 [14]3 years ago
3 0

Answer:

the bus traveled 26.25 miles

Step-by-step explanation:

hello

the speed ​​is a physical quantity that expresses the relationship between the space traveled by an object, the time taken for it ,and it is given by

step 1

v=\frac{d}{t}

where, d is the distance traveled and t is the time it takes

now, if we isolate d

v=\frac{d}{t}\\d=v*t\\

step 2

convert 45 minutes into hours

60 min =1 hour

45 min= x?

x=45 h/60

x=0.75 hours

step 3

replace data

d=35 mph*0.75 hours\\d=26.25\ miles

the bus traveled 26.25 miles

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Revenue from Monorail Service, Las Vegas In 2005 the Las Vegas monorail charged $3 per ride and had an average ridership of abou
Karolina [17]

Answer:

A)The required linear demand equation ( q ) = -4500p + 41500

B) $4.61

   $95680.55

C) No it would not have been possible by charging a suitable price

Step-by-step explanation:

<u>A)  find the linear demand equation</u>

given two points ; ( 3, 28000 ) and ( 5, 19000 )

slope ( m ) = ( y2 - y1 ) / ( x2 - x1 )

                 = ( 19000 - 28000 ) / ( 5 - 3 )  = -4500

slope intercept is represented as ; y = mx + b

where y( 28000) = -4500(3) + b

  hence b = 41500  

hence ; y = -4500x + 41500

The required linear demand equation ( q ) = -4500p + 41500   ----- ( 1 )

p = price per ride

<u>B ) Determine the price the company should charge to maximize revenue from ridership  and corresponding daily revenue</u>

Total revenue ( R ) = qp

                               = p ( -4500p + 41500 )

  hence R = -4500p^2 + 41500p  ------ ( 2 )

To determine the price that should maximize revenue from ridership we will equate R = -4500p^2 + 41500p  to a quadratic equation R(p) = ap^2 + bp + c

where a = -4500 ,  b = 41500 , c = 0

hence p = -\frac{b}{2a}  = - \frac{41500}{2(-4500)} =  4.61

$4.61 is the price the company should charge to maximize revenue from ridership

corresponding daily revenue = R = -4500p^2 + 41500 p

where p = $4.61

hence R = -4500(4.61 )^2 + 41500(4.61) = $95680.55

C) No it would not have been possible by charging a suitable price

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3 years ago
There are 18 oranges in box A and 36 in box B, what is the ratio of box A to box B
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Answer:

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Step-by-step explanation:

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box A : box B

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telo118 [61]

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Step-by-step explanation:

7 0
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Which of the following gives an example of a set that is closed under addition.
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I believe the answer is D
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Estimate the area of Alberta in square miles. Show your reasoning
ella [17]

Answer:  About 278,250\ mi^2

Step-by-step explanation:

The missing figure is attached.

Notice in the first picture that Alberta has a complex shape.

You can calculate the area of a complex shape by decomposing it into polygons whose areas can be calculated easily.

Observe the second picture. Notice that it can be descompose into two polygons: A trapezoid and a rectangle.

The area of the trapezoid  can be calcualted with the formula:

A_t=\frac{h}{2}(B+b)

Where "h" is the height, "B" is the long base and "b" is short base.

And the area of the rectangle can be found with the formula:

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Wkere "l" is the lenght and "w" is the width.

Then, the apprximate area of Alberta is:

A=\frac{h}{2}(B+b)+lw

Substituting vallues, you get:

A=(\frac{(410\ mi-180\ mi)}{2})(760\ mi+470\ mi)+(180\ mi)(760\ im)\\\\A=141,450\ mi^2+136,800\ mi^2\\\\A=278,250\ mi^2

Therefore, the area of of Alberta is about 278,250\ mi^2.

5 0
3 years ago
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