Identify the horizontal asymptote of f(x) = quantity 2 x minus 1 over quantity x squared minus 7 x plus 3.
1 answer:
We must recall that a horizontal asymptote is the value/s of y that the given function approaches to but never reaches. To find this in a rational function, we compare the expressions with highest degree in the numerator and denominator. There are three possible outcome when this happens.
1. if the highest degree (highest exponent) in the numerator is bigger than that of the denominator, then there won't be any horizontal asymptote.
2. if the highest degree in the denominator is bigger, then the horizontal symptote would be y = 0.
3. if they have the same highest degree, then we just get the quotient of their coefficient.
Now, going back to our function, we have
From this we can see that the highest degree in the numerator is 1 (from 2x) and 2 (from x²) for the denominator. Clearly, it shows that its denominator has a higher degree. And from our discussion, we can conclude that the horizontal asymptote would be y = 0.
Answer: y = 0
1. if the highest degree (highest exponent) in the numerator is bigger than that of the denominator, then there won't be any horizontal asymptote.
2. if the highest degree in the denominator is bigger, then the horizontal symptote would be y = 0.
3. if they have the same highest degree, then we just get the quotient of their coefficient.
Now, going back to our function, we have
From this we can see that the highest degree in the numerator is 1 (from 2x) and 2 (from x²) for the denominator. Clearly, it shows that its denominator has a higher degree. And from our discussion, we can conclude that the horizontal asymptote would be y = 0.
Answer: y = 0
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Answer: (0, 0), (-3, -9), (3, -9)
which means that "g of a number" is equal to "negative, the square of that number".
According to the function we calculate g(0), g(-3), g(3) as follows:
This means that we have the points (0, 0), (-3, -9), and (3, -9) lying on the graph of the function.
Answer: (0, 0), (-3, -9), (3, -9)
Answer:
We conclude that the mean IQ of her students is different from the national average.
Step-by-step explanation:
We are given that the national mean (μ) IQ score from an IQ test is 100 with a standard deviation (s) of 15.
The dean of a college want to test whether the mean IQ of her students is different from the national average. For this, she administers IQ tests to her 144 students and calculates a mean score of 113
Let, Null Hypothesis, :
= 100 {means that the mean IQ of her students is same as of national average}
Alternate Hypothesis, :
100 {means that the mean IQ of her students is different from the national average}
(a) The test statistics that will be used here is One sample z-test statistics;
T.S. = ~ N(0,1)
where, Xbar = sample mean score = 113
s = population standard deviation = 15
n = sample of students = 144
So, test statistics =
= 10.4
Now, at 0.05 significance level, the z table gives critical value of 1.96. Since our test statistics is more than the critical value of z which means our test statistics will fall in the rejection region and we have sufficient evidence to reject our null hypothesis.
Therefore, we conclude that the mean IQ of her students is different from the national average.