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wariber [46]
3 years ago
7

If a pair of standard dice are rolled, what is the probability that one of the dice will be a 3 and the other a 4?

Mathematics
1 answer:
34kurt3 years ago
6 0
The probability for any number to be rolled on a dice is 1/6. So the probability that you would roll a 3 is 1/6 and the probability that you would roll a 4 is 1/6. You would then multiple time together to get your final probability 1/6 x 1/6 = 0.0278 or 2.78% probability.
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When two chords intersect in a cirlce, the ___ of the segments of one chord?
barxatty [35]
<span>Whenever two chords intersect in a circle, the products of their segments are equal.</span> 

Source:
http://www.1728.org/circangl.htm


3 0
3 years ago
two hundred tickets for the school play were sold. tickets cost $2 for students and $3 for adults. the total amount collected wa
Gre4nikov [31]
A system of equations is good for a problem like this.
Let x be the number of student tickets sold
Let y be the number of adult tickets sold
x + y = 200
2x + 3y = 490
x = 200 - y
2(200 - y) + 3y = 490
400 - 2y + 3y = 490
400 + y = 490
y = 90
The number of adult tickets sold was 90.
x + 90 = 200 --> x = 110
2x + 3(90) = 490 --> 2x + 270 = 490 --> 2x = 220 --> x = 110
The number student tickets sold was 110.
4 0
3 years ago
Simplify the expression to show 7m+4n-m-2n​
Valentin [98]

Answer:

6m + 2n

Step-by-step explanation:

we have to added up the like therms

like therms are therms that have the same literal part

6m + 2n

5 0
3 years ago
Read 2 more answers
Let theta be an angle in quadrant II such that cos theta = -2/3
Iteru [2.4K]

Answer:

So we have \csc(\theta)=\frac{3 \sqrt{5}}{5} \text{ and } \tan(\theta)=\frac{-\sqrt{5}}{2}.

Step-by-step explanation:

Ok so we are in quadrant 2, that means sine is positive while cosine is negative.

We are given \cos(\theta)=\frac{-2}{3}(\frac{\text{adjacent}}{\text{hypotenuse}}).

So to find the opposite we will just use the Pythagorean Theorem.

a^2+b^2=c^2

(2)^2+b^2=(3)^2

4+b^2=9

b^2=5

b=\sqrt{5}  This is the opposite side.

Now to find \csc(\theta) and \tan(\theta).

\csc(\theta)=\frac{\text{hypotenuse}}{\text{opposite}}=\frac{3}{\sqrt{5}}.

Some teachers do not like the radical on bottom so we will rationalize the denominator by multiplying the numerator and denominator by sqrt(5).

So \csc(\theta)=\frac{3}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}=\frac{3 \sqrt{5}}{5}.

And now \tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}=\frac{\sqrt{5}}{-2}=\frac{-\sqrt{5}}{2}.

So we have \csc(\theta)=\frac{3 \sqrt{5}}{5} \text{ and } \tan(\theta)=\frac{-\sqrt{5}}{2}.

7 0
3 years ago
Read 2 more answers
Solve the equation 0.5p - 3.45 = -1.2
Ivenika [448]
<span>0.5p - 3.45 = -1.2 ⇒ p=4.5</span>
3 0
4 years ago
Read 2 more answers
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