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alina1380 [7]
4 years ago
11

It is believed that students who begin studying for final exams a week before the test score differently than students who wait

until the night before. Suppose you want to test specifically that students who study earlier have an average score that is less than the average score for students who wait to study. What are the hypotheses for this test if early studiers are group 1 and procrastinators are group 2
Mathematics
1 answer:
olga55 [171]4 years ago
7 0

Answer:

H0 : μ1 = μ2

H0 : μ1 < μ2

Step-by-step explanation:

Given :

Early studiers, group 1, mean μ1

Late studiers, group 2 ; mean, μ2

To test the claim that students who study earlier have an average score that is less than the average score for students

The null hypothesis ; there is no difference in average score

H0 : μ1 = μ2

Alternative hypothesis is early studiers means is less than average score of late studiers

H0 : μ1 < μ2

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The length of the hypotenuse in a 30°-60°-90° triangle is 6√10yd. What is the
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a) The polynomial in expanded form is f(x) = x^{3}-2\cdot x^{2}-21\cdot x -18.

b) The slant asymptote is represented by the linear function is y = -x + 1.

c) There is a discontinuity at x = 2  with a slant asymptote.

a) In this question we are going to use the Factor Theorem, which establishes that polynomial are the result of products of binomials of the form x-r_{i}, where r_{i} is the i-th root of the polynomial and the grade is equal to the quantity of roots. Therefore, the polynomial f(x) has the following form:

f(x) = (x-6)\cdot (x+1)\cdot (x+3)

And the expanded form is obtained by some algebraic handling:

f(x) = (x-6)\cdot (x^{2}+4\cdot x +3)

f(x) = x\cdot (x^{2}+4\cdot x + 3)-6\cdot (x^{2}+4\cdot x +3)

f(x) = x^{3}+4\cdot x^{2}+3\cdot x -6\cdot x^{2}-24\cdot x -18

f(x) = x^{3}-2\cdot x^{2}-21\cdot x -18 (1)

The polynomial in expanded form is f(x) = x^{3}-2\cdot x^{2}-21\cdot x -18.

b) In this question we divide the polynomial found in a) (in factor form) by the polynomial x^{2}-x -2 (also in factor form). That is:

g(x) = \frac{(x-6)\cdot (x+1)\cdot (x+3)}{(x-2)\cdot (x+1)}

g(x) = \frac{(x-6)\cdot (x+3)}{x-2} (2)

The slant asymptote is defined by linear function, whose slope (m) and intercept (b) are determined by the following expressions:

m =  \lim_{x \to \pm \infty} \frac{g(x)}{x} (3)

b =  \lim_{x \to \pm \infty} [g(x)-x] (4)

If g(x) = \frac{(x-6)\cdot (x+3)}{x-2}, then the equation of the slant asymptote is:

m =  \lim_{x \to 2} \frac{(x-6)\cdot (x+3)}{x\cdot (x-2)}

m =  \lim_{x \to \pm \infty} \left(\frac{x^{2}-3\cdot x -18}{x^{2}-2\cdot x} \right)

m =  1

b =  \lim_{x \to \pm \infty} \left(\frac{x^{2}-3\cdot x -18}{x-2}-x \right)

b =  \lim_{x \to \pm \infty} \left(\frac{x^{2}-3\cdot x - 18-x^{2}+2\cdot x}{x-2}\right)

b =  \lim_{n \to \infty} \left(\frac{-x-18}{x-2} \right)

b = -1

The slant asymptote is represented by the linear function is y = -x + 1.

c) The number of discontinuities in rational functions is equal to the number of binomials in the denominator, which was determined in b). Hence, we have a discontinuity at x = 2  with a slant asymptote.

We kindly invite to check this question on asymptotes: brainly.com/question/4084552

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