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artcher [175]
3 years ago
7

Question 2 (Multiple Choice Worth 5 points)

Mathematics
1 answer:
steposvetlana [31]3 years ago
6 0
Number of remaining rose cards = 6
number of remaining cards = 8
P(rose card) = 6/8

P(blue marble) = 18/50
P(tail) = 30/50
P(blue marble and tail) = 18/50 x 30/50 = 540/2500
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Need help with AP CAL
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.

5 0
2 years ago
0 points<br> 6a – 14a2 + 8a + 1 by a - 2
Mekhanik [1.2K]

Answer:

-14a -14- 27/a-2 hope this helps

Step-by-step explanation:

6 0
3 years ago
The larger of two numbers is five more than twice the smaller number. The sum of the two numbers is 38. Find the numbers.
Damm [24]

Answer:

I would set this up as:

let: lesser number = x

     greater number = 2x + 5

Restating the word problem:

lesser number + greater number = 38

        x           +  2x + 5             = 38

Solving:  3x + 5 = 38

                  - 5     -5

             ---------------

              3x      =  33

              ---          ---

               3            3

               x        =  11

Substitution for the greater number:

        2(11) + 5

          22    + 5

              27

8 0
3 years ago
Read 2 more answers
5x + 7x + 2x + 8 -2 please helppp
Murljashka [212]

Answer:

= 14X + 6

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
A population has the following characteristics. (a) A total of 75% of the population survives the first year. Of that 75%, 25% s
Artemon [7]

Answer:

= \left[\begin{array}{ccc}1344\\84\\28\end{array}\right]  \left \begin{array}{ccc}{0 \  \leq  age   \leq  1 }\\{ 1 \  \leq  age   \leq  2 }\\{2 \  \leq  age  \leq 3}\end{array}\right

i.e after the first year ;

there 1344 members in the first age class

84 members for the second age class; and

28 members for the third age class

Step-by-step explanation:

We can deduce that the age distribution vector x represents the number of population members for each age class; Given that in each class of age there are 112 members present.

The current age distribution vector is as follows:

x = \left[\begin{array}{ccc}1&1&2\\1&1&2\\1&1&2\end{array}\right] \left[\begin{array}{ccc}{0 \  \leq  age   \leq  1 }\\{ 0 \  \leq  age   \leq  2 }\\{0 \  \leq  age   \leq 3}\end{array}\right]

Also , the age transition matrix is as follows:

L = \left[\begin{array}{ccc}3&6&3\\0.75&0&0 \\0&0.25&0\end{array}\right]

After 1 year ; the age distribution vector will be :

x_2 =Lx_1 = \left[\begin{array}{ccc}3&6&3\\0.75&0&0 \\0&0.25&0\end{array}\right]  \left[\begin{array}{ccc}1&1&2\\1&1&2\\1&1&2\end{array}\right]

= \left[\begin{array}{ccc}1344\\84\\28\end{array}\right]  \left \begin{array}{ccc}{0 \  \leq  age   \leq 1 }\\{ 1 \  \leq  age   \leq  2 }\\{2 \  \leq  age   \leq  3}\end{array}\right

6 0
3 years ago
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