If X is a continuous random variable having the Uniform probability distribution between 1 and 5, what is the probability that, of six observations taken of this random variable, three of them will be larger than or equal to 4?

And that would represent the probability of success p =0.25.

Step-by-step explanation:

For this case we know that X represent a continuous random variable and we know that the distribution for X is given by:

And we can begin finsding the following probability for an individual case:

We can use the complement rule and we have this:

And we can use the cumulative distribution function for the continuous random variable given by:

And replacing we got:

And that would represent the probability of success p =0.25.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let Y the random variable of interest, on this case we now that:

The probability mass function for the Binomial distribution is given as:

Where (nCy) means combinatory and it's given by this formula:

And we want to find this probability:

answer: 26

step-by-step explanation:

well, michelle wanted to measure the height of her school's flagpole she placed a mirror on the ground 48 feet from the flagpole then walk backwards until she was able to see the top of the pole in the mirror her eyes were 5 feet above the ground and she was 12 ft from the mirror using similar triangles find the height of the flagpole to the nearest tenth of a foot find the geometric mean of them pair of numbers. its easy math

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.