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3 years ago

If U = {all positive integers) and A = {x|x EU and x is an odd positive integer}, which describes AC?

Mathematics

2 answers:

Galina-37 [17]3 years ago

7 0

Answer:

last choice

Step-by-step explanation:

If we are looking at positive integers and we take out the odd ones, the only one's that are left are the even ones.

So last choice. (Also I assume AC meant the complement of A)

Helen [10]3 years ago

6 0

Answer:

Step-by-step explanation:

ed2020

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3 years ago

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Prove that the triangle EDF is isosceles. Give reasons for your answer.

Gekata [30.6K]

Answer:

$\triangle EDF$ is isosceles.

Step-by-step explanation:

Please have a look at the attached figure.

We are given the following things:

$\angle EDF = y$

$\text{External }\angle DFG = 90 +\dfrac{y}{2}$

Let us try to find out $\angle E$ and $\angle DFE$ . After that we will compare them.

Finding  $\angle DFE$ :

Side EG is a straight line so $\angle GFE = 180$

$\angle GFE$ is sum of internal $\angle DFE$ and external $\angle DFG$

$\angle GFE = 180 = \angle DFE + \angle DFG\\\Rightarrow 180 = \angle DFE + (90+\dfrac{y}{2})\\\Rightarrow \angle DFE = 180 - 90 - \dfrac{y}{2}\\\Rightarrow \angle DFE = 90 - \dfrac{y}{2} ....... (1)$

Finding  $\angle E$ :

Property of external angle: External angle in a triangle is equal to the sum of two opposite internal angles of a triangle.

i.e. external $\angle DFG$ = $\angle E + \angle EDF$

$\Rightarrow 90+\dfrac{y}{2} = \angle E + y\\\Rightarrow \angle E = 90+\dfrac{y}{2} -y\\\Rightarrow \angle E = 90-\dfrac{y}{2} ....... (2)$

Comparing equations (1) and (2):

It can be clearly seen that:

$\angle DFE = \angle E =90-\dfrac{y}{2}$

The two angles of $\triangle EDF$ are equal hence $\triangle EDF$ is isosceles.

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2 years ago

A graph has minutes on the x-axis and ounces of popcorn on the y-axis. A curve goes through points (0, 0), (50, 12), and (110, 1

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Answer:

No :)

Step-by-step explanation:

the graph does not make a straight line. :)

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2 years ago

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Find the probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if:a) a

lara31 [8.8K]

Answer:

A) 0.0009765625

B) 0.0060466176

C) 2.7756 x 10^(-17)

Step-by-step explanation:

A) This problem follows a binomial distribution. The number of successes among a fixed number of trials is; n = 10

If a 0 bit and 1 bit are equally likely, then the probability to select in 1 bit is; p = 1/2 = 0.5

Now the definition of binomial probability is given by;

P(K = x) = C(n, k)•p^(k)•(1 - p)^(n - k)

Now, we want the definition of this probability at k = 10.

Thus;

P(x = 10) = C(10,10)•0.5^(10)•(1 - 0.5)^(10 - 10)

P(x = 10) = 0.0009765625

B) here we are given that p = 0.6 while n remains 10 and k = 10

Thus;

P(x = 10) = C(10,10)•0.6^(10)•(1 - 0.6)^(10 - 10)

P(x=10) = 0.0060466176

C) we are given that;

P((x_i) = 1) = 1/(2^(i))

Where i = 1,2,3.....,n

Now, the probability for the different bits is independent, so we can use multiplication rule for independent events which gives;

P(x = 10) = P((x_1) = 1)•P((x_2) = 1)•P((x_3) = 1)••P((x_4) = 1)•P((x_5) = 1)•P((x_6) = 1)•P((x_7) = 1)•P((x_8) = 1)•P((x_9) = 1)•P((x_10) = 1)

This gives;

P(x = 10) = [1/(2^(1))]•[1/(2^(2))]•[1/(2^(3))]•[1/(2^(4))]....•[1/(2^(10))]

This gives;

P(x = 10) = [1/(2^(55))]

P(x = 10) = 2.7756 x 10^(-17)

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