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Katen [24]
3 years ago
6

Solve the problem and then click on the correct graph. y < |x + 3|

Mathematics
2 answers:
trapecia [35]3 years ago
6 0
The second picture should be the correct graph or the one which has its blue on the bottom side.
Marianna [84]3 years ago
3 0
The 2nd graph is your answer!
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Please help! Identify the asymptotes of the graph
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In this situation, the asymptopes are the dotted lines.


Looking at the graph, the equations of such lines are

X=2, Y=1

7 0
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(4 Points)
bearhunter [10]

Answer:

Step-by-step explanation:

You know the lengths of two sides and the angle between them. Use the Law of Cosines to solve the triangle.

PR^2 = 6^2 + 9^2 - 2*6*9*cos(120°)

= 36 + 81 - 108(-0.5)

= 13

5 0
3 years ago
A baseball team won 20 games and lost 10 games. What percent of the games did the team win? 33% 50% 66% 200%
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The answer is C. 66.6% is one third

Step-by-step explanation:

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Benicio rolls a pair of number cubes. What is the probability that the sum is 6 if the number are different? What is the probabi
jok3333 [9.3K]

Answer:

The probabilities are \dfrac{2}{15} and \dfrac{4}{5}.

Step-by-step explanation:

When a dice is rolled, then total possible outcomes = 6 (i.e. 1,2,3,4,5 or 6).

So, when a pair of dice is rolled,

Total possible outcomes, n(S) = 6 × 6 = 36

Consider A represents the event of getting a sum of 6 and B represents the event of different numbers appear.

A={(1,5),(2,4),(3,3),(4,2),(5,1)}

B={(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5)}

A∩B= {(1,5),(2,4),(4,2),(5,1)}

So, n(A) = 5, n(B) = 30, n(A∩B) = 4.

\text{Probability}=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}

Thus,

P(A)=\frac{n(A)}{n(S)}=\frac{5}{36}

P(B)=\frac{n(B)}{n(S)}=\frac{30}{36}=\frac{5}{6}

P(A\cap B)=\frac{n(A\cap B)}{n(S)}=\frac{4}{36}=\frac{1}{9}

Hence, the probability that the sum is 6 if the number are different,

P(\frac{A}{B})=\frac{P(A\cap B)}{P(B)}

         =\frac{\frac{1}{9}}{\frac{5}{6}}

         =\frac{6}{45}

         =\frac{2}{15}

Similarly,

The probability that the numbers are different if the sum is 6,

P(\frac{B}{A})=\frac{P(A\cap B)}{P(A)}

         =\frac{\frac{1}{9}}{\frac{5}{36}}

         =\frac{36}{45}

         =\frac{4}{5}

8 0
3 years ago
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