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Olegator [25]
3 years ago
6

The average and standard deviation of a measurement is 15.0 ± 5.0. if 100 students measured the value, how many would you expect

to get a value greater than 20?
Mathematics
1 answer:
AURORKA [14]3 years ago
4 0

Solution: We are given:

\mu=15, \sigma=5

Using the empirical rule, we have:

\mu \pm \sigma covers 68% of data.

Also the percentage of values below mean = Percentage of values above mean = 50%

Now, let's find the z score for x=20

z=\frac{20-15}{5}=1

Therefore, the percentage of values greater than 1 standard deviation above mean 50\% - \frac{68\%}{2} =50\%-34\%=16%

Expected number of students = 16% of 100 = 16


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The Chang family is on their way home from a cross-country road trip. During the trip, the function D(t)=3260−55t can be used to
nadya68 [22]

Answer:

D(12) = 2,600 miles

It means a distance of 2,600 miles is already traveled from home after 12 hours

Step-by-step explanation:

To find D(12); all we have to do is to substitute the value of 12 for D

We have this as;

D(12) = 3260-55(12)

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In the context of this problem, what this mean is that the distance away from home is 2,600 miles after traveling 12 hours

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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

5 0
2 years ago
=
Kitty [74]

Answer:

2

Step-by-step explanation:

6 0
2 years ago
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