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KatRina [158]
3 years ago
13

Using the information given, select the statement that can deduce the line segments to be parallel. If there are none, then sele

ct none.
When m<2 = m<6

Mathematics
1 answer:
Nataly [62]3 years ago
4 0
The second choice is correct.  Given that angle 2 and angle 6 are congruent, the largest in each of the corresponding triangles are vertical angles meaning that two known angles are known in each triangle (and they are equal) which means that the last "unknown" (angles 3 and 7) are congruent.  The lines DC and AB correspond with these defined, congruent triangles meaning they are parallel.
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If the odds against Deborahs winning first prize in a chess tournament are 1 to 11 what is the probability of the event that she
Lemur [1.5K]

Answer: \dfrac{11}{12}

Step-by-step explanation:

Given

The odds against winning in a chess tournament are 1 to 11.

Odds is defined as the ratio of the probability of occurrence to the non-occurrence of event.

\therefore \text{Probability that event will occur is P'=}\dfrac{1}{1+11}\\\\\Rightarrow P'=\dfrac{1}{12}

Probability of non-occurrence i.e. she wins the first prize is

\Rightarrow P=1-\dfrac{1}{12}\\\\\Rightarrow P=\dfrac{11}{12}

7 0
3 years ago
What is the least common denominator of 1/9 and 2/3
Kipish [7]

Answer:

9

Step-by-step explanation:

All u have to do is think only about two denominators, 9 and 3

And then memorize the multiplication of any number out of these two, until u get a number which is divisible by the other number. Its like,

9×1= 9

i

is it divisible by 3?

9/3= 3

Oh yes, it is divisible by 3 with no remainings. So that is the least common denominator

8 0
3 years ago
Marshall has two pieces of string, one 12 feet long and the other 18 feet long. For an art project, he wants to cut them up to p
lukranit [14]
For the first piece of string that is 12 feet long, he could make two 6 foot pieces. For the second piece of string that is 18 feet long he could make two 9 foot pieces. Hope I could help! :D
4 0
3 years ago
What is the domain of the square root function graphed below?
OLEGan [10]
The answer is X >= 0
3 0
3 years ago
A certain Bookstore that sells A Million books wants to hire you, and you get the job if you can answer these problems correctly
Maslowich

Answer:

a) P ( X = 2 ) = 0.23028

b) P ( X < 4 ) = 0.95689

c) P ( X ≥ 3 | X ≥ 2 ) = 0.38292

Step-by-step explanation:

Given:-

- The parameter for the poisson distribution is given, λ = 1.3.

- Declare a random variable (X) which is the number of books sold in the next minute:

                               X ~ Po (1.3)

Find:-

a) What's the probability that the store will sell 2 books in the next minute? b) What's the probability that the store will sell less than 4 books in the next minute? c) What's the probability that the store will sell at least 3 books in the next minute given that it sells at least 2 books in the next minute?

Solution:-

a) The required probability P ( X = 2 ). Can be computed by using the pmf for the poisson distribution:

                       P(X = x ) =\frac{ (lambda)^k e^(^-^l^a^m^b^d^a^)}{k!}\\\\P(X = x )  =\frac{ (1.3)^k e^(^-^1^.^3^)}{k!}

Where, "k" is the number of books sold in next minute.

- Now compute P ( X = 2 ) :

                       P(X = 2 )  =\frac{ (1.3)^2 e^(^-^1^.^3^)}{2!}\\\\P(X = 2 )  = 0.23028    

b) The required probability P ( X < 4 ). Can be computed by using the pmf for the poisson distribution and summing individual terms from 0 - 3:

                      P(X < 4 )  = P ( X = 0) + P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 )\\\\P(X < 4 )  = \frac{ (1.3)^0 e^(^-^1^.^3^)}{0!}+ \frac{ (1.3)^1 e^(^-^1^.^3^)}{1!}+ \frac{ (1.3)^2 e^(^-^1^.^3^)}{2!} + \frac{ (1.3)^3 e^(^-^1^.^3^)}{3!}\\\\P(X < 4 )  = 0.27253 + 0.35429 + 0.23028 + 0.09979 = 0.95689

c) The required probability P ( X ≥ 3 | X ≥ 2 ). We have to consider the conditional probability as follows:

                   P ( X ≥ 3 | X ≥ 2 ) = P ( X ≥ 3 & X ≥ 2 ) / P (X ≥ 2 )

                                               = P ( X ≥ 3 ) / P (X ≥ 2 )

                                               = P ( X > 2 ) / P ( X > 1 )

                                               = [ 1 - P ( X ≤ 2 ) ] / [ 1 - P ( X ≤ 1 ) ]  

- Compute P ( X ≤ 2 ) & P ( X ≤ 1 ) using pmf:

                     P ( X ≤ 2 ) = 0.27253 + 0.35429 + 0.23028

                                      = 0.8571

                     P ( X ≤ 1 ) = 0.27253 + 0.35429

                                      = 0.62682

- Use the expression developed for conditional probability to evaluate the required probability:

                     P ( X ≥ 3 | X ≥ 2 ) = [ 1 - P ( X ≤ 2 ) ] / [ 1 - P ( X ≤ 1 ) ]

                                                  = [ 1 - 0.8571 ] / [ 1 - 0.62682 ]

                                                  = 0.38292

8 0
3 years ago
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