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inna [77]
3 years ago
13

(x)=\frac{3x-6}{5-2x}" align="absmiddle" class="latex-formula">
Domain:
V.A:
Roots:
Y-Int:
H.A:
Holes:
O.A:

Also draw on the graph.

Mathematics
1 answer:
Svetradugi [14.3K]3 years ago
3 0

i) The given function is

f(x)=\frac{3x-6}{5-2x}

The domain is all real values except the ones that will make the denominator zero.

5-2x=0

-2x=-5

x=2.5

The domain is all real values except, x=2.5.

ii) To find the vertical asymptote, we equate the denominator to zero and solve for x.

5-2x=0

-2x=-5

x=2.5

iii) If we equate the numerator to zero, we get;

3x-6=0

3x=6

This implies that;

x=2

iv) To find the y-intercept, we put x=0 into the given function to get;

f(0)=\frac{3(0)-6}{5-2(0)}.

f(0)=\frac{-6}{5}.

f(0)=-\frac{6}{5}.

v)

The degrees of both numerator and the denominator are the same.

The ratio of the coefficient of the degree of the numerator to that of the denominator will give us the asymptote.

The horizontal asymptote  is y=-\frac{3}{2}.

vi) The function has no common factors that are at least linear.

The function has no holes in it.

vii) This rational function has no oblique asymptotes because it is a proper rational function.

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3 years ago
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It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute
NemiM [27]

Answer:

a) 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

b) 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mildly obese

Normally distributed with mean 373 minutes and standard deviation 67 minutes. So \mu = 373, \sigma = 67

A) What is the probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes?

So n = 5, s = \frac{67}{\sqrt{5}} = 29.96

This probability is 1 subtracted by the pvalue of Z when X = 410.

Z = \frac{X - \mu}{s}

Z = \frac{410 - 373}{29.96}

Z = 1.23

Z = 1.23 has a pvalue of 0.8907.

So there is a 1-0.8907 = 0.1093 = 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

Lean

Normally distributed with mean 526 minutes and standard deviation 107 minutes. So \mu = 526, \sigma = 107

B) What is the probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes?

So n = 5, s = \frac{107}{\sqrt{5}} = 47.86

This probability is 1 subtracted by the pvalue of Z when X = 410.

Z = \frac{X - \mu}{s}

Z = \frac{410 - 526}{47.86}

Z = -2.42

Z = -2.42 has a pvalue of 0.0078.

So there is a 1-0.0078 = 0.9922 = 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.

7 0
3 years ago
What is the oblique asymptote of the function f(x) = the quantity x squared minus 5x plus 6 over the quantity x minus 4?
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Answer:

x - 1

Step-by-step explanation:

We know that, a slant or oblique asymptote of a rational function is the asymptote that helps in determining the direction of the function.

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Now, we divide the numerator by denominator using long division method and the first two terms in the quotient ( forming a linear function ) is the equation of the oblique asymptote.

We are given the rational function, f(x) = \frac{x^{2}-5x+6}{x-4}.

After dividing we get that, the quotient is x - 1.

Hence, the equation of the oblique asymptote is x-1.

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We have 4(x^2 - 2x + 4) - 7 = 4x^2 - 8x + 16 - 7 = 4x^2 - 8x + 9;
So, 4x^2 - 8x + 9 = <span>ax^2+bx+c ;
The value of b is -8.</span>
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Someone asked the same question here before, check it out

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