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3 years ago

How can I do this with the function (x+2)/x-3

1 answer:

3 0

Y= (x+2)/(x-3)
This function exists for x≠3 (value that render the denominator nil)
If x →3 , then y→ ∞, then:
Vertical Asymptote x=3

Horizontal asymptotes ↔ lim y = 1
x→∞
Horizontal asymptotes y =1

x-intercept: for y= 0, x = -2

y-intercept: for x= 0, y = -2/3

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For parts a and bâ, use technology to estimate the following. âa) The critical value of t for a 90â% confidence interval with df

rjkz [21]

Answer:

a) For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

$t_{\alpha/2} =\pm 2.35$

b) For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

$t_{\alpha/2} =\pm 2.62$

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

Part a

For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

$t_{\alpha/2} =\pm 2.35$

Part b

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

$t_{\alpha/2} =\pm 2.62$

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3 years ago

Why is 5 a rational number?

NNADVOKAT [17]

Every integer is a rational number, since each integer n can be written in the form n/1. For example 5 = 5/1 and thus 5 is a rational number. i hope this helps you

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3 years ago

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antoniya [11.8K]

Answer:

The first blank is BAL and FGH

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explanation

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AlekseyPX

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