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Gwar [14]
3 years ago
5

The correlation coefficient for blood pressure and amount of vegetables eaten in a group of people is −0.7. Analyze the followin

g statement:
High blood pressure is caused by not eating vegetables.

Is this a reasonable conclusion?
Mathematics
2 answers:
vfiekz [6]3 years ago
6 0
The correlation coefficient for blood pressure and amount of vegetables eaten which is a negative number (-0.7) suggests that they are in inverse relation. Eating more vegetable should lower the blood pressure. In the same manner, not eating vegetables would likely cause high blood pressure. 
nasty-shy [4]3 years ago
5 0

Answer:

No; even though there is a strong negative correlation, not eating vegetables doesn't necessarily cause high blood pressure

Step-by-step explanation:

just took the quiz

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Identify the domain and range
UkoKoshka [18]

Answer:

Domain: 1 ≤ x ≤ 4

Range : 1 ≤ f(x) ≤ 4

Step-by-step explanation:

The domain of a function f(x) is the limit within which the values of x varies.

Here, in the graph, it shows that the maximum value of x is 4 and the minimum value of x is 1.

Therefore, the domain of the function is 1 ≤ x ≤ 4

Again the range of a function f(x) is the limit within which the values of f(x) vary.

Here, the graph shows that the maximum value of f(x) is 4 and the minimum value of f(x) is 1.

Therefore, the range of the function f(x) is 1 ≤ f(x) ≤ 4. (Answer)

6 0
3 years ago
If alpha and beta are the zeroes of the polynomial 6x2+x-2 find the value og alpha/beta + beta/alpha
castortr0y [4]
6x^2+x-2=6x^2+4x-3x-2=2x(3x+2)-1(3x+2)\\\\=(3x+2)(2x-1)\\\\3x+2=0\to x=-\frac{2}{3}\\\\2x-1=0\to x=\frac{1}{2}\\\\\alpha=-\frac{2}{3};\ \beta=\frac{1}{2}\\\\\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{-\frac{2}{3}}{\frac{1}{2}}+\frac{\frac{1}{2}}{-\frac{2}{3}}=-\frac{2}{3}\cdot\frac{2}{1}-\frac{1}{2}\cdot\frac{3}{2}=-\frac{4}{3}-\frac{3}{4}\\\\=-\frac{16}{12}-\frac{9}{12}=-\frac{25}{12}=-2\frac{1}{12}


use\ Vieta's\ formula:\\\\\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{\alpha^2+2\alpha\beta+\beta^2-2\alpha\beta}{\alpha\beta}=\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}=\frac{(\alpha+\beta)^2}{\alpha\beta}-2\\\\\alpha+\beta=\frac{-b}{a};\ \alpha\beta=\frac{c}{a}\\\\\frac{(\alpha+\beta)^2}{\alpha\beta}-2=\frac{\left(\frac{-b}{a}\right)^2}{\frac{c}{a}}-2=\frac{b^2}{a^2}\cdot\frac{a}{c}-2=\frac{b^2}{ac}-2

6x^2+x-2\\\\a=6;\ b=1;\ c=-2\\\\\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{1^2}{6\cdot(-2)}-2=\frac{1}{-12}-2=-2\frac{1}{12}
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3 years ago
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8 0
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