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Ksenya-84 [330]
3 years ago
14

The admission to a local carnival is $8.25 per person and $1.50 for each ride. How much more will it cost a group of 3 friends t

o ride 8 rides each than to ride 3 rides each?
$7.50
$22.50
$32.25
$48.75
Mathematics
2 answers:
beks73 [17]3 years ago
8 0
3*(8-3)*(1.50)=22.5 second option
Anit [1.1K]3 years ago
7 0

Answer:

$22.50

Step-by-step explanation:

Number of friends = 3

Admission cost per person = $8.25

So, total admission cost for 3 persons =3 \times 8.25

                                                                =24.75

Each friend takes 8 rides

So, total rides taken by 3 friends = 8 \times 3 =24

Cost of 1 ride = $1.50

So, Cost of 24 rides =  24 \times 1.50 =36

So, total cost for group of 3 friends in which each friend takes 8 rides = 24.75+36 =$ 60.75

If Each friend takes 3 rides

So, total rides taken by 3 friends = 3 \times 3 =9

Cost of 1 ride = $1.50

So, Cost of 9 rides =  9 \times 1.50 =13.5

So, total cost for group of 3 friends in which each friend takes 3 rides = 24.75+13.5 =$ 38.25

Difference in both the costs = $60.75 -$38.25=$22.50

Hence it will cost $22.50 more than a group of 3 friends to ride 8 rides each than to ride 3 rides each.

So, Option B $22.50 is correct.

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anzhelika [568]

Answer: Choice C

\displaystyle \frac{1}{2}\left(1 - \frac{1}{e^2}\right)

============================================================

Explanation:

The graph is shown below. The base of the 3D solid is the blue region. It spans from x = 0 to x = 1. It's also above the x axis, and below the curve y = e^{-x}

Think of the blue region as the floor of this weirdly shaped 3D room.

We're told that the cross sections are perpendicular to the x axis and each cross section is a square. The side length of each square is e^{-x} where 0 < x < 1

Let's compute the area of each general cross section.

\text{area} = (\text{side})^2\\\\\text{area} = (e^{-x})^2\\\\\text{area} = e^{-2x}\\\\

We'll be integrating infinitely many of these infinitely thin square slabs to find the volume of the 3D shape. Think of it like stacking concrete blocks together, except the blocks are side by side (instead of on top of each other). Or you can think of it like a row of square books of varying sizes. The books are very very thin.

This is what we want to compute

\displaystyle \int_{0}^{1}e^{-2x}dx\\\\

Apply a u-substitution

u = -2x

du/dx = -2

du = -2dx

dx = du/(-2)

dx = -0.5du

Also, don't forget to change the limits of integration

  • If x = 0, then u = -2x = -2(0) = 0
  • If x = 1, then u = -2x = -2(1) = -2

This means,

\displaystyle \int_{0}^{1}e^{-2x}dx = \int_{0}^{-2}e^{u}(-0.5du) = 0.5\int_{-2}^{0}e^{u}du\\\\\\

I used the rule that \displaystyle \int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx which says swapping the limits of integration will have us swap the sign out front.

--------

Furthermore,

\displaystyle 0.5\int_{-2}^{0}e^{u}du = \frac{1}{2}\left[e^u+C\right]_{-2}^{0}\\\\\\= \frac{1}{2}\left[(e^0+C)-(e^{-2}+C)\right]\\\\\\= \frac{1}{2}\left[1 - \frac{1}{e^2}\right]

In short,

\displaystyle \int_{0}^{1}e^{-2x}dx = \frac{1}{2}\left[1 - \frac{1}{e^2}\right]

This points us to choice C as the final answer.

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