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3 years ago

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A box has 1 red marble, 1 blue marble, and 4 green marbles of the same shape and size. Dan draws a marble randomly from the box,

replaces it, and then draws another marble randomly. What is the probability of drawing two green marbles in a row? Explain your answer. PLEASSSEEEE HELP! IM DESPERATE

1 answer:

lora16 [44]3 years ago

8 0

Answer:

2/3 or 66.66..%

Step-by-step explanation:

First, add all the number of marbles and you get 6. Second you just divide the number of green marbles which is 4 by 6 and you get 2/3 or if they want it in percent then, 66.66..%

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part 2. Find the value of the trig function indicated, use for that Pythagorean theorem to find the third side if you need it.

andreev551 [17]

Answer: $\bold{5)\ \cos \theta=\dfrac{\sqrt{11}}{6}\qquad 6)\ \tan \theta =\dfrac{8}{17}\qquad 7)\ \cos \theta = \dfrac{4}{3}\qquad 8)\ \cos \theta = \dfrac{\sqrt{10}}{10}}$

<u>Step-by-Step Explanation:</u>

Pythagorean Theorem is: a² + b² = c² , <em>where "c" is the hypotenuse</em>

$5)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{3\sqrt{11}}{18}\quad \rightarrow \large\boxed{\dfrac{\sqrt{11}}{6}}$

Note: (15)² + (3√11)² = hypotenuse² → hypotenuse = 18

$6)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{8}{17}\quad =\large\boxed{\dfrac{8}{17}}$

Note: 8² + 15² = hypotenuse² → hypotenuse = 17

$7)\ \tan \theta=\dfrac{\text{side opposite to}\ \theta}{\text{side adjacent to}\ \theta}=\dfrac{20}{15}\quad \rightarrow \large\boxed{\dfrac{4}{3}}$

Note: hypotenuse not needed for tan

$8)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{2}{2\sqrt{10}}\quad =\large\boxed{\dfrac{\sqrt{10}}{10}}$

Note: 2² + 6² = hypotenuse² → hypotenuse = 2√10

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2 years ago

PLZ HEP WITH ONE QUESTION

grin007 [14]

There's 2/8 yellow ones, that simplifies to 1/4

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2 years ago

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What are the factors of the mathematical expression 24p-27p²

Vlad [161]

Answer:

1expn=3p

2expn=8-9p

3p(8-9p)

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Solve the system of equations.<br><br><br> <br><br> −5y−10x=45<br> −3y+10x=−5<br>

Fynjy0 [20]

(-2,-5)
The first goal with equations like these is to get one of the variables to go away. You can do that by adding/subtracting one equation from the other. Then you solve for the one variable. Then you plug that answer into one of the original equations and solve.

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Given: ABC is a right triangle with right angle C. AC=15 centimeters and m∠A=40∘ . What is BC ? Enter your answer, rounded to th

konstantin123 [22]

In order to answer this question, the figure in the first picture will be helpful to understand what a right triangle is. Here, a right angle refers to $90\°$ .

However, if we want to solve the problem we have to know certain things before:

In the second figure is shown a general right triangle with its three sides and another given angle, we will name it $\alpha$ :

The side <u>opposite to the right angle</u> is called The Hypotenuse (h)

The side <u>opposite to the angle $\alpha$ </u> is called the Opposite (O)

The side <u>next to the angle $\alpha$ </u> is called the Adjacent (A)

So, going back to the triangle of our question (first figure):

The Hypotenuse is AB

The Opposite is BC

The Adjacent is AC

Now, if we want to find the length of each side of a right triangle, we have to use the <u>Pythagorean Theorem</u> and T<u>rigonometric Functions:</u>

Pythagorean Theorem

$h^{2}=A^{2} +O^{2}$ (1)

Trigonometric Functions (here are shown three of them):

Sine: $sin(\alpha)=\frac{O}{h}$ (2)

Cosine: $cos(\alpha)=\frac{A}{h}$ (3)

Tangent: $tan(\alpha)=\frac{O}{A}$ (4)

In this case the function that works for this problem is cosine (3), let’s apply it here:

$cos(40\°)=\frac{AC}{h}$

$cos(40\°)=\frac{15}{h}$ (5)

And we will use the Pythagorean Theorem to find the hypotenuse, as well:

$h^{2}=AC^{2}+BC^{2}$

$h^{2}=15^{2}+BC^{2}$ (6)

$h=\sqrt{225+BC^2}$ (7)

Substitute (7) in (5):

$cos(40\°)=\frac{15}{\sqrt{225+BC^2}}$

Then clear BC, which is the side we want:

${\sqrt{225+BC^2}}=\frac{15}{cos(40\°)}$

${{\sqrt{225+BC^2}}^2={(\frac{15}{cos(40\°)})}^2$

$225+BC^{2}=\frac{225}{{(cos(40\°))}^2}$

$BC^2=\frac{225}{{(cos(40\°))}^2}-225$

$BC=\sqrt{158,41}$

$BC=12.58$

Finally $BC$ is approximately 13 cm

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