A yield sign measures 30 inches on all three sides. what is the area of the sign

Mathematics

1 answer:

Nookie1986 [14]3 years ago

5 0

Answer:

450

Step-by-step explanation:

To find an area of the triangle, you must first multiply the base and the height.

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Determine whether the given procedure results in a binomial distribution. If it is not binomial, identify the requirements that

masha68 [24]

Answer:

This procedure does not result in a binomial distribution because in each trial there are more than two possible outcomes, that is, this problem cannot be simplified as a yes or no question.

Step-by-step explanation:

For the binomial probability distribution being satisfied, two things are necessary:

1) The answer can only be two outcomes, two possibilities, for example, yes/no questions.

2) The probability of a success(an answer) on each trial must be the same. That is, if we are going to survey 100 people if they are Buffalo Bills fans, they will answer yes or no, and each of those 100 people will have the same probability of being a Buffalo Bills fans.

In this problem, we have that:

Surveying 20 college students and recording their favorite TV show.

There is infinite possibilities for each answer. One can answer Narcos, other Monday Night Football, other Sportscentre, and so it goes.

So this procedure does not result in a binomial distribution because in each trial there are more than two possible answers, that is, this problem cannot be simplified as a yes or no question.

If for example, we asked the 20 students if Monday Night Football was their favorite show, then they could answer only yes or no, so it would be a binomial procedure.

3 0

3 years ago

Y=arccos(1/x)<br><br> Please help me do them all! I don’t know derivatives :(

Stells [14]

Answer:

$f(x) = {sec}^{ - 1} x \\ let \: y = {sec}^{ - 1} x \rightarrow \: x = sec \: y\\ \frac{dx}{dx} = \frac{d(sec \: y)}{dx} \\ 1 = \frac{d(sec \: y)}{dx} \times \frac{dy}{dy} \\ 1 = \frac{d(sec \: y)}{dy} \times \frac{dy}{dx} \\1 = tan \: y.sec \: y. \frac{dy}{dx} \\ \frac{dy}{dx} = \frac{1}{tan \: y.sec \: y} \\ \frac{dy}{dx} = \frac{1}{ \sqrt{( {sec}^{2} \: y - 1}) .sec \: y} \\ \frac{dy}{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \frac{d( {sec}^{ - 1}5x) }{dx} = \frac{1}{ |5x | \sqrt{25 {x }^{2} - 1} }\\\\y=arccos(\frac{1}{x})\Rightarrow cosy=\frac{1}{x}\\x=secy\Rightarrow y=arcsecx\\\therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} }$