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Eduardwww [97]
3 years ago
12

Jerry thinks that x < 4 is a solution to the following inequality: 2x + 10 < 4x - 20

Mathematics
1 answer:
I am Lyosha [343]3 years ago
7 0

Answer:

x>15

Step-by-step explanation:

Given the inequality: 2x + 10 < 4x - 20

Collect like terms

2x-4x<-20-10

-2x<-30

x > -30/-2

x>15

Hence the solution is x>15

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Explanation given below.

Step-by-step explanation:

The first step is to put the parabola in the form  ax^2+bx+c , which is the <em>standard form of a parabola</em>

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<u>Note:</u> a is the coefficient before x^2 term, b is the coefficient before x term, and c is the independent constant term

The axis of symmetry divides the parabola symmetrically. The axis of symmetry has the equation  x=-\frac{b}{2a}

Where <em><u>a and b are the respective values shown above</u></em>

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Step-by-step explanation:

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A survey said that 3 out of 5 students enrolled in higher education took at least one online course last fall. Explain your calc
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Answer:

a) 60% probability that student took at least one online course

b) 40% probability that student did not take an online course

c) 12.96% probability that all 4 students selected took online courses.

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they took at least one online course last fall, or they did not. The probability of a student taking an online course is independent of 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.

3 out of 5 students enrolled in higher education took at least one online course last fall.

This means that p = \frac{3}{5} = 0.6

a) If you were to pick at random 1 student enrolled in higher education, what is the probability that student took at least one online course?

This is P(X = 1) when n = 1. So

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

P(X = 1) = C_{1,1}.(0.6)^{1}.(0.4)^{0} = 0.6

60% probability that student took at least one online course.

b) If you were to pick at random 1 student enrolled in higher education, what is the probability that student did not take an online course?

This is P(X = 0) when n = 1.

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

P(X = 0) = C_{1,1}.(0.6)^{0}.(0.4)^{1} = 0.4

40% probability that student did not take an online course

c) Now, consider the scenario that you are going to select random select 4 students enrolled in higher education. Find the probability that all 4 students selected took online courses

This is P(X = 4) when n = 4.

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

P(X = 4) = C_{4,4}.(0.6)^{4}.(0.4)^{0} = 0.1296

12.96% probability that all 4 students selected took online courses.

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