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Daniel [21]
3 years ago
13

I NEED HELP PLEASE !!!!

Mathematics
1 answer:
Softa [21]3 years ago
4 0

Answer:

See below

Step-by-step explanation:

So we have the following six values:

77.65488139\\\sqrt{0.156}\\368. 5468432...\\253,897,615.\bar3\\14.19274128...\\\\\sqrt{\frac{4}{9} }

And we want to determine which are irrational.

Recall what irrational numbers, are they are numbers that do not<em> </em>repeat nor terminate. In other words, they never end, and their digits never repeat.

So, let's go through each one.

77.65488139 is rational. Yes, it has a lot of decimal places but notice there are no ellipses after it. Thus, it terminates and is rational.

√(0.156) is about (use a calculator) 0.39496835... It is not repeating nor does it end. Thus, √(0.156) is irrational.

368.5468432... has the ellipses. Thus, it doesn't repeat and doesn't terminate. It's irrational.

253,897,615.3 has a bar over the 3 at the end. This means that this repeats. Therefore, while it doesn't terminate, it is rational.

14.19274128... has an ellipses. It doesn't repeat nor terminate. It's irrational.

Lastly, the square root can be simplified to 2/3. This is approximately .6666... While this doesn't terminate, it repeats, so it's rational.

Edit: Grammar

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jeka94

5(x-8)-(x+6)=18
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not sure if this is correct but i hope it helps!!!
7 0
3 years ago
Are your finances, buying habits, medical records, and phone calls really private? A real concern for many adults is that comput
Andrej [43]

Answer:

a)There is a 4.88% probability that none is concerned that employers are monitoring phone calls.

b)There is a 7.89% probability that all are concerned that employers are monitoring phone calls.

c)There is a 37.23% probability that exactly two are concerned that employers are monitoring phone calls.

Step-by-step explanation:

The binomial probability is the probability of exactly x successes on n repeated trials in an experiment which has two possible outcomes (commonly called a binomial experiment).

It is given by the following formula:

P = C_{n,x}.p^{n}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of a success.

In this problem, a success is being concerned that employers are monitoring phone calls.

53% of adults are concerned that employers are monitoring phone calls, so p = 0.53

(a) Out of four adults, none is concerned that employers are monitoring phone calls.

Four adults, so n = 4.

Is the probability of 0 successes, so x = 0.

P = C_{n,x}.p^{n}.(1-p)^{n-x}

P = C_{4,0}.(0.53)^{0}.(0.47)^{4}

P = 0.0488

There is a 4.88% probability that none is concerned that employers are monitoring phone calls.

(b) Out of four adults, all are concerned that employers are monitoring phone calls.

Four adults, so n = 4.

Is the probability of 4 successes, so x = 4.

P = C_{n,x}.p^{n}.(1-p)^{n-x}

P = C_{4,0}.(0.53)^{4}.(0.47)^{0}

P = 0.0789

There is a 7.89% probability that all are concerned that employers are monitoring phone calls.

(c) Out of four adults, exactly two are concerned that employers are monitoring phone calls.

Four adults, so n = 4.

Is the probability of 4 successes, so x = 2.

P = C_{n,x}.p^{n}.(1-p)^{n-x}

P = C_{4,2}.(0.53)^{2}.(0.47)^{2}

P = 0.3723

There is a 37.23% probability that exactly two are concerned that employers are monitoring phone calls.

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Step-by-step explanation:

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