Express 35 1/3 in simplest radical form.

The graph of a rational function f is shown below.

Mashutka [201]

We have y+2 = 0 and x - 2 = 0. The provided function has an x and y-intercept of -2 and +2, respectively. There is no vertical asymptote. Two is the horizontal asymptote.

<h3>What is a graph?</h3>

A diagram depicting the relationship between two or more variables, each measured along with one of a pair of axes at right angles.

The y-intercept of a function is determined by the intersection of its graph with the y-axis. The value of y on the y-axis at which the considered function crosses it is called the y-intercept.

Assume the following equation: y = f (x)

We have x =0- 2 and y+2 = 0,The x and y intercept of the given function is -2 and +2.

The vertical asymptote is none. The horizontal asymptote is 2.

Hence,we have y+2 = 0 and x - 2 = 0. The provided function has an x and y-intercept of -2 and +2, respectively. There is no vertical asymptote. Two is the horizontal asymptote.

To learn more about the graph, refer to the link;

brainly.com/question/14375099

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4 0

1 year ago

A survey was conducted to determine the average age at which college seniors hope to retire in a simple random sample of 101 sen

tatyana61 [14]

Answer:

96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].

Step-by-step explanation:

We are given that a survey was conducted to determine the average age at which college seniors hope to retire in a simple random sample of 101 seniors, 55 was the average desired retirement age, with a standard deviation of 3.4 years.

Firstly, the Pivotal quantity for 96% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average desired retirement age = 55 years

$\sigma$ = sample standard deviation = 3.4 years

n = sample of seniors = 101

$\mu$ = true mean retirement age of all college students

<em>Here for constructing 96% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.</em>

<u>So, 96% confidence interval for the population mean, </u> $\mu$ <u> is ;</u>

P(-2.114 < $t_1_0_0$ < 2.114) = 0.96 {As the critical value of t at 100 degree