Graph the image of the polygon after a reflection in the line y=x<br><br> PLEASE HELP

Problem 2.2.4 Your Starburst candy has 12 pieces, three pieces of each of four flavors: berry, lemon, orange, and cherry, arrang

kkurt [141]

Answer:

a) P=0

b) P=0.164

c) P=0.145

Step-by-step explanation:

We have 12 pieces, with 3 of each of the 4 flavors.

You draw the first 4 pieces.

a) The probability of getting all of the same flavor is 0, because there are only 3 pieces of each flavor. Once you get the 3 of the same flavor, there are only the other flavors remaining.

b) The probability of all 4 being from different flavor can be calculated as the multiplication of 4 probabilities.

The first probability is for the first draw, and has a value of 1, as any flavor will be ok.

The second probability corresponds to drawing the second candy and getting a different flavor. There are 2 pieces of the flavor from draw 1, and 9 from the other flavors, so this probability is 9/(9+2)=9/11≈0.82.

The third probability is getting in the third draw a different flavor from the previos two draws. We have left 10 candys and 4 are from the flavor we already picked. Then the third probabilty is 6/10=0.6.

The fourth probability is getting the last flavor. There are 9 candies left and only 3 are of the flavor that hasn't been picked yet. Then, the probability is 3/9=0.33.

Then, the probabilty of picking the 4 from different flavors is:

$P=1\cdot\dfrac{9}{11}\cdot\dfrac{6}{10}\cdot\dfrac{3}{9}=\dfrac{162}{990}\approx0.164$

c) We can repeat the method for the previous probabilty.

The first draw has a probability of 1 because any flavor is ok.

In the second draw, we may get the same flavor, with probability 2/11, or we can get a second flavor with probability 9/11. These two branches are ok.

For the third draw, if we have gotten 2 of the same flavor (P=2/11), we have to get a different flavor (we can not have 3 of the same flavor). This happen with probability 9/10.

If we have gotten two diffente flavors, there are left 4 candies of the picked flavors in the remaining 10 candies, so we have a probabilty of 4/10.

For the fourth draw, independently of the three draws, there are only 2 candies left that satisfy the condition, so we have a probability of 2/9.

For the first path, where we pick 2 candies of the same flavor first and 2 candies of the same flavor last, we have two versions, one for each flavor, so we multiply this probability by a factor of 2.

We have then the probabilty as:

$P=2\cdot\left(1\cdot\dfrac{2}{11}\right)\cdot\left(\dfrac{9}{10}\cdot\dfrac{2}{9}\right)+\left(1\cdot\dfrac{9}{11}\cdot\dfrac{4}{10}\cdot\dfrac{2}{9}\right)\\\\\\P=2\cdot\dfrac{36}{990}+\dfrac{72}{990}=\dfrac{144}{990}\approx0.145$

5 0

2 years ago

Maggie claims that there are transformations that preserve the length of the rectangle's sides. Which of the following transform

Nataly_w [17]

Answer:

B, D, and E

Step-by-step explanation:

6 0

1 year ago

gia got some books from the library yesterday. she got five books after returning three. Gia now has six books. how many library

larisa [96]

Gia began with four books

5 0

1 year ago

Read 2 more answers

A train is traveling at a constant speed of 105 miles per hour. How many feet does it travel in 4 seconds? Remember that 1 mile

Alex73 [517]

9240ft per second

(105mi/1hr * 5280ft/1mi * 1hr/60sec)

5 0

1 year ago

The angle \theta_1θ

enyata [817]

Answer:

$sin\theta_1 = \dfrac{\sqrt{217}}{19}$

Step-by-step explanation:

It is given that:

$cos\theta_1 = -\dfrac{12}{19}$

And we have to find the value of $sin\theta_1 = ?$

As per trigonometric identities, the relation between $sin\theta\ and \ cos\theta$ can represented as:

$sin^2\theta + cos^2\theta = 1$

Putting $\theta_1$ in place of $\theta$ Because we are given

$sin^2\theta_1 + cos^2\theta_1 = 1$

Putting value of cosine:

$cos\theta_1 = -\dfrac{12}{19}$

$sin^2\theta_1 + (\dfrac{12}{19})^2 = 1\\\Rightarrow sin^2\theta_1 + \dfrac{144}{361} = 1\\\Rightarrow sin^2\theta_1 = 1-\dfrac{144}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{361-144}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{217}{361}\\\Rightarrow sin\theta_1 = +\sqrt{\dfrac{217}{361}}, -\sqrt{\dfrac{217}{361}}\\\Rightarrow sin\theta_1 = +\dfrac{\sqrt{217}}{19}, -\dfrac{\sqrt{217}}{19}$

It is given that $\theta_1$ is in 2nd quadrant and value of sine is always positive in 2nd quadrant. So, the answer is.

$\Rightarrow sin\theta_1 = \dfrac{\sqrt{217}}{19}$

8 0

1 year ago

Graph the image of the polygon after a reflection in the line y=x PLEASE HELP