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neonofarm [45]
3 years ago
12

In a University of Wisconsin (UW) study about alcohol abuse among students, 100 of the 40,858 members of the student body in Mad

ison were sampled and asked to complete a questionnaire. One question asked was, "On how many days in the past week did you consume at least one alcoholic drink?" a. Identify the population and the sample. b. For the 40,858 students at UW, one characteristic of interest was the percentage who would respond "zero" to this question. For the 100 students sampled, suppose 29% gave this response. Does this mean that 29% of the entire population of UW students would make this response? Explain.
Mathematics
1 answer:
dimaraw [331]3 years ago
7 0

Answer:

a) The population is 40,858 students and the sample is 100.

b) No

Step-by-step explanation:

a) The population would be the 40,858 members of the student body. Since we are only applying the questionnaire to 100 students, the sample would be 100.

b) 29% of the students answered "zero" to the question on how many days in the past week they consumed at least one alcoholic drink. This means that 29 out of 100 students gave this answer. However, this doesn't mean that 29% of the entire population of UW would give this response. Why is that? Because our sample is very small so it might not be representative of the whole population. Equally, the results from such a sample cannot be exactly the same results we would get from an entire population.

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Please help!! What is the solution to the quadratic inequality? 6x2≥10+11x
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Answer:

The solution of the inequation 6\cdot x^{2} \geq 10 + 11\cdot x is \left(-\infty,-\frac{2}{3}\right]\cup\left[\frac{5}{2},+\infty\right).

Step-by-step explanation:

First of all, let simplify and factorize the resulting polynomial:

6\cdot x^{2} \geq 10 + 11\cdot x

6\cdot x^{2}-11\cdot x -10 \geq 0

6\cdot \left(x^{2}-\frac{11}{6}\cdot x -\frac{10}{6} \right)\geq 0

Roots are found by Quadratic Formula:

r_{1,2} = \frac{\left[-\left(-\frac{11}{6}\right)\pm \sqrt{\left(-\frac{11}{6} \right)^{2}-4\cdot (1)\cdot \left(-\frac{10}{6} \right)} \right]}{2\cdot (1)}

r_{1} = \frac{5}{2} and r_{2} = -\frac{2}{3}

Then, the factorized form of the inequation is:

6\cdot \left(x-\frac{5}{2}\right)\cdot \left(x+\frac{2}{3} \right)\geq 0

By Real Algebra, there are two condition that fulfill the inequation:

a) x-\frac{5}{2} \geq 0 \,\wedge\,x+\frac{2}{3}\geq 0

x \geq \frac{5}{2}\,\wedge\,x \geq-\frac{2}{3}

x \geq \frac{5}{2}

b) x-\frac{5}{2} \leq 0 \,\wedge\,x+\frac{2}{3}\leq 0

x \leq \frac{5}{2}\,\wedge\,x\leq-\frac{2}{3}

x\leq -\frac{2}{3}

The solution of the inequation 6\cdot x^{2} \geq 10 + 11\cdot x is \left(-\infty,-\frac{2}{3}\right]\cup\left[\frac{5}{2},+\infty\right).

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