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katovenus [111]
4 years ago
6

Which graph is the solution to lxl > 10?

Mathematics
1 answer:
ale4655 [162]4 years ago
5 0

\huge\text{Hey there!}

\huge\text{Note}\rightarrow \leq \& \geq \ = \huge\text{closed circle}\\ < \& > = \huge\text{(o)(p)(e)(n)(e)(d) circle}

\huge\text{So, we know that A or C could be our answer}.

\huge\text{Eliminate option B and D}

→ \huge\text{Find the absolute value of the problem}

→ \huge\text{You can now know that  it could be:}  

\huge\text{x}\large\huge{\ >}\huge\text{ 10 or x }< \huge\text{ -10}

→ \huge\text{Possible answer \#1: x }>\huge\text{10}\\\\\huge\text{Possible answer \#2: x}

→ \boxed{\boxed{\huge\text{Answer: C}}}  

\huge\text{Option A. would be too small} \huge\checkmark

\text{Good luck on your assignment and enjoy your day!}

~\frak{LoveYourselfFirst:)}

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At a local University, it is reported that 81% of students own a wireless device. If 5 students are selected at random, what is
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34.87% probability that all 5 have a wireless device

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they own a wireless device, or they do not. The probability of a student owning a wireless device is independent from other students. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which C_{n,x} is the number of different combinations 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 X happening.

81% of students own a wireless device.

This means that p = 0.81

If 5 students are selected at random, what is the probability that all 5 have a wireless device?

This is P(X = 5) when n = 5. So

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

P(X = 5) = C_{5,5}.(0.81)^{5}.(0.19)^{0} = 0.3487

34.87% probability that all 5 have a wireless device

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3 years ago
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