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3 years ago

What is the range of y= sec-'(x)? PreCal, send help please!!

Mathematics

1 answer:

beks73 [17]3 years ago

6 0

Given:

The function is:

$y=\sec^{-1}(x)$

To find:

The range of the given function.

Solution:

We have,

$y=\sec^{-1}(x)$

The range of secant inverse function is:

$Range=\{y|0\leq y\leq \pi , y\neq \dfrac{\pi}{2}\}$

The range of the given function in interval notation is:

$Range=\left[0,\dfrac{\pi}{2}\right)\text{ and }\left( \dfrac{\pi}{2}, \pi\right ]$

Therefore, the correct option is C.

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3 years ago

Assume that the Poisson distribution applies and that the mean number of hurricanes in a certain area is 6.9 per year. a. Find t

VLD [36.1K]

Answer:

a) 9.52% probability that, in a year, there will be 4 hurricanes.

b) 4.284 years are expected to have 4 hurricanes.

c) The value of 4 is very close to the expected value of 4.284, so the Poisson distribution works well here.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

6.9 per year.

This means that $\mu = 6.9$

a. Find the probability that, in a year, there will be 4 hurricanes.

This is P(X = 4).

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 4) = \frac{e^{-6.9}*(6.9)^{4}}{(4)!}$

$P(X = 4) = 0.0952$

9.52% probability that, in a year, there will be 4 hurricanes.

b. In a 45-year period, how many years are expected to have 4 hurricanes?

For each year, the probability is 0.0952.

Multiplying by 45

45*0.0952 = 4.284.

4.284 years are expected to have 4 hurricanes.

c. How does the result from part (b) compare to a recent period of 45 years in which 4 years had 4 hurricanes? Does the Poisson distribution work well here?

The value of 4 is very close to the expected value of 4.284, so the Poisson distribution works well here.

5 0

2 years ago

Delicious Candy markets a two-pound box of assorted chocolates. Because of imperfec- tions in the candy making equipment, the ac

saveliy_v [14]

Answer:

The required probability is 0.533.

Step-by-step explanation:

Consider the provided information.

The actual weight of the chocolate has a uniform distribution ranging from 31 to 32.5 ounces.

Let x is the random variable for the actual weight of chocolate.

According to PDF function.

$P(a\leq x\leq b)=\int\limits^b_a {f(x)} \, dx$

Where $f(x)=\left\{\begin{matrix}\frac{1}{b-a} & a$

It is given that ranging from 31 to 32.5 ounces.

Substitute a=31 and b=32.5 in above function.

$f(x)=\left\{\begin{matrix}\frac{1}{32.5-31} & 31$

$f(x)=\left\{\begin{matrix}\frac{1}{1.5} & 31$

We need to find the probability that a box weighs less than 31.8 ounces

Now according to PDF:

$P(x$

Hence, the required probability is 0.533.

6 0

3 years ago

The diagram below shows circle with radii OA

Finger [1]

Answer:

Length of arc AB = 4π

Step-by-step explanation:

Formula to get the arc length is given by,

Length of arc = $\frac{\theta}{360}(2\pi r)$

Here, θ = central angle subtended by the arc AB

r = Radius of the circle

By substituting the values in the formula,

Therefore, length of arc AB = $\frac{120}{360}(2\pi )(6)$

= 40π

3 0

3 years ago