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3 years ago

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Assume that each bag contains 5 balls. Bag a contains 2 red and 3 white, while bag b contains 1 red, 1 white, and 3 blue. You ra

ndomly select one ball from bag a, note the color, and place the ball in bag b. You then select a ball from bag b at random and make note of its color. What is the probability that both balls are red?

1 answer:

Tcecarenko [31]3 years ago

7 0

Answer:

2/15

Step-by-step explanation:

Firstly, we calculate the probability of picking a red ball from bag A.

P = number of expected outcomes/number of total outcomes.

P(R1) in bag A = 2/5

Now, we take a ball from bag A and place in bag B. If we selected a red ball, the number of balls in bag B now equals = 2 red balls while the total number of balls now equal 6 balls. Now, the probability is :

P(R2) = 2/6 = 1/3.

The total probability P(R) = P(R1) × P(R2) = 2/5 × 1/3 = 2/15

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Q8: These Triangles are similar. What is the value of X?

Umnica [9.8K]

Answer:

The answer to your question is x = 15

Step-by-step explanation:

Use proportions to solve this problem

$\frac{x}{20} = \frac{12}{16}$

Solve for "x"

x = $\frac{12(20)}{16}$

Simplification

x = $\frac{240}{16}$

Result

x = 15

3 0

3 years ago

Match the circle equations in general form with their corresponding equations in standard form. Not all will be used.

Xelga [282]

<span>The standard form of the equation of a circumference is given by the following expression:

</span> $(x-h)^{2}+(y-k)^{2}=r^{2} \\ \\ where \ (h, k) \ is \ the \ center \ of \ the \ circumference \ and \ r \ the \ radius$
<span>
On the other hand, the general form is given as follows:

</span> $x^{2}+y^{2}+Dx+Ey+F=0 \\ \\ where: \\ D=-2h, \ E=-2k, \ F=h^{2}+k^{2}-r^{2}$ <span>

In this way, we can order the mentioned equations as follows:

Equations in Standard Form:

</span> $\bold{a)} \ (x-6)^{2}+(y-4)^{2}=56 \\ \bold{b)} \ (x-2)^{2} + (y+6)^{2}=60 \\ \bold{c)} \ (x+2)^{2}+(y+3)^{2}=18 \\ \bold{d)} \ (x+1)^{2}+(y-6)^{2}=46$

Equations in General Form:

$\bold{1)} \ x^{2}+y^{2}-4x+12y-20=0 \\ \bold{2)} \ x^{2}+y^{2}+6x-8y-10=0 \\ \bold{3)} \ 3x^{2}+3y^{2}+12x+18y-15=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 3, \ the \ equation \ becomes: \\ x^{2}+y^{2}+4x+6y-5=0 \\ \\ \bold{4)} \ 5x^{2}+5y^{2}-10x+20y-30=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 5, \ the \ equation \ becomes: \\ x^{2}+y^{2}-2x+4y-6=0 \\ \\ \bold{5)} \ 2x^{2}+2y^{2}-24x-16y-8=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 2, \ the \ equation \ becomes: \\ x^{2}+y^{2}-12x-8y-4=0$

$\bold{6)} \ x^{2}+y^{2}+2x-12y$

So let's match each equation:

$\bold{From \ a)} \\ \\ (h,k)=(6,4),\ r=2\sqrt{14} \\ D=-12, \ E=-8 \\ F=-4$

Then, its general form is:

$x^{2}+y^{2}-12x-8y-4=0$

<em><u>First. a) matches 5) </u></em>

$\bold{From \ b)} \\ \\ (h,k)=(2,-6),\ r=2\sqrt{15} \\ D=-4, \ E=12 \\ F=-20$

Then, its general form is:

$x^{2}+y^{2}-4x+12y-20=0$

<em><u>Second. b) matches 1) </u></em>

$\bold{From \ c)} \\ \\ (h,k)=(-2,-3),\ r=3\sqrt{2} \\ D=4, \ E=6 \\ F=-5$

Then, its general form is:

$x^{2}+y^{2}+4x+6y-5=0$

<em><u>Third. c) matches 3)</u></em>

$\bold{From \ d)} \\ \\ (h,k)=(-1,6),\ r=\sqrt{46} \\ D=2, \ E=-12, \ F=-9$

Then, its general form is: $x^{2}+y^{2}+2x-12y-9=0$

<em><u>Fourth. d) matches 6)</u></em>

6 0

3 years ago

Read 2 more answers

What is the area of the pentagon

vlada-n [284]

The area of the pentagon is 140

3 0

3 years ago

Read 2 more answers

The set of ordered pairs represents a

Murrr4er [49]

Answer:

The rule that represents the function is $y=x^2+1$ therefore the function is $f(x)=x^2+1$

Step-by-step explanation:

We have 5 ordered pairs in the plane xy. This means that <em>every pair has the form (x, y).</em>

Then, we have 5 values of x, which will give us 5 values of y, using the rule that represents the function.

<u>The easy evaluation is that when x=0, the value of y is y=1,</u> and then we can evaluate the rule for x=-1, and x=1, <em>the value of y is the same, y=2</em>. We can see here that we have a parabolic function, that is not centered in the origin of coordinates because when x=0, y=1.

So <u>we propose the rule </u> $y=x^2+1$ <u> which is correct for the first 3 values of x.</u>

Now, <em>we evaluate the proposed rule when x=2, and when x=3</em>. This evaluations can be written as

$f(2)=2^2+1=5$

$f(3)=3^2+1=10$

Therefore, the rule is correct, and the function is

$f(x)=x^2+1$

8 0

3 years ago

PLZ HELP! Need this ASAP! Thx so much :)

saul85 [17]

they are giving you the answer AB=AC, so what that means is that the angles are congruent to each other so if angle AC=15 and angle DC=5 all the angles will be the same.

5 0

3 years ago