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2 years ago

Which of the following is true about horizontal asymptotes?

1 answer:

8 0

A function has a horizontal asymptote at the value of y = a if the line y = a can be used to estimate the end behavior of a function and if f ( x ) → a as x → ∞ or x → − ∞ is the correct statement about horizontal asymptotes. Option A

<h3>What are horizontal asymptotes?</h3>

A horizontal asymptote of a graph can be defined as a horizontal line at y = b where the graph tend to approach the line as an inputs approach to infinity ( ∞ or –∞).

A slant asymptote of a graph is known as a slanted line y = mx + b where the graph approaches the line as the inputs approach the positive infinity ∞ or to the infinity –∞.

Thus, a function has a horizontal asymptote at the value of y = a if the line y = a can be used to estimate the end behavior of a function and if f ( x ) → a as x → ∞ or x → − ∞ is the correct statement about horizontal asymptotes. Option A

Learn more about horizontal asymptotes here:

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Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). To estimate

Leviafan [203]

Answer:

Probability of having student's score between 505 and 515 is 0.36

Given that z-scores are rounded to two decimals using Standard Normal Distribution Table

Step-by-step explanation:

As we know from normal distribution: z(x) = (x - Mu)/SD

where x = targeted value; Mu = Mean of Normal Distribution; SD = Standard Deviation of Normal Distribution

Therefore using given data: Mu (Mean) = 510, SD = 10.4 we have z(x) by using z(x) = (x - Mu)/SD as under:

In our case, we have x = 505 & 515

Approach 1 using Standard Normal Distribution Table:

z for x=505: z(505) = (505-510)/10.4 gives us z(505) = -0.48

z for x=515: z(515) = (515-510)/10.4 gives us z(515) = 0.48

Afterwards using Normal Distribution Tables and rounding the values to two decimals we find the probabilities as under:

P(505) using z(505) = 0.32

Similarly we have:

P(515) using z(515) = 0.68

Now we may find the probability of student's score between 505 and 515 using:

P(505 < x < 515) = P(515)-P(505) = 0.68 - 0.32 = 0.36

PS: The standard normal distribution table is being attached for reference.

Approach 2 using Excel or Google Sheets:

P(x) = norm.dist(x,Mean,SD,Commutative)

P(505) = norm.dist(505,510,10.4,1)

P(515) = norm.dist(515,510,10.4,1)

Probability of student's score between 505 and 515= P(515) - P(505) = 0.36

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6 0

3 years ago

6415 divided 32 estimate

dsp73

Answer:

200

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

There are 27 students in Ms. Langley's

Yuliya22 [10]

Answer:

405 Problems

Step-by-step explanation:

One student in class has 15 problems.

There are 27 students in the class.

Let's <u>multiply</u> 15 by 27.

$15*27=405$

<em>In all, there would be 405 problems assigned to the whole class.</em>

8 0

3 years ago

Figures that have the same size and same shape. Their corresponding angles and sides are the same size. whats the answer?

Irina-Kira [14]

Answer:

prostokąt i kwadrat

7 0

3 years ago

what is the probability of seeing a sample mean for 21 observations less or equal to the sample mean that we observed

kramer

Answer:

$P(x \le 21) = 0.69146$

Step-by-step explanation:

The missing parameters are:

$n = 64$ --- population

$\mu = 20$ --- population mean

$\sigma = 16$ -- population standard deviation

Required

$P(x \le 21)$

First, calculate the sample standard deviation

$\sigma_x = \frac{\sigma}{\sqrt n}$

$\sigma_x = \frac{16}{\sqrt {64}}$

$\sigma_x = \frac{16}{8}$

$\sigma_x = 2$

Next, calculate the sample mean $\bar_x$

$\bar x = \mu$

So:

$\bar x = 20$

So, we have:

$\sigma_x = 2$

$\bar x = 20$

$x = 21$

Calculate the z score

$x = \frac{x - \mu}{\sigma}$

$x = \frac{21 - 20}{2}$

$x = \frac{1}{2}$

$x = 0.50$

So, we have:

$P(x \le 21) = P(z \le 0.50)$

From the z table

$P(z \le 0.50) = 0.69146$

So:

$P(x \le 21) = 0.69146$

4 0

3 years ago