The piecewise function f(x) is defined below. What is the value of f(–2)?

Mathematics

2 answers:

nikdorinn [45]3 years ago

7 0

Answer:

D. $f(-2)=3$

Step by step explanation:

We have been given a piece wise function:

$f{x}=\left\{\begin{matrix}-x^{2}; &x< -2 \\3;&-2\leq x< 0\\x+2; & x\geq 0\end{matrix}\right$

We are asked to find value of f(-2).

In our piece function when we will input values of x less than -2( $x$ ) , it will give us output as $-x^{2}$

For values greater than or equal to 0 our function will give output as x+2.

We can see from our piece wise function that when x is greater than or equal to -2 and less than 0 our function gives a constant output, that is 3.

Therefore, $f(-2)=3$ and option D is the correct choice.

Usimov [2.4K]3 years ago

3 0

$f(-2)=3$ , because $f(x)=3$ for $-2\leq x\ \textless \ 0$

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Suppose that the distance, in miles, that people are willing to commute to work is an exponential random variable with a decay p

garik1379 [7]

Answer:

m = $\frac{1}{20}$
μ = 20
σ = 20

The probability that a person is willing to commute more than 25 miles is 0.2865.

Step-by-step explanation:

Exponential probability distribution is used to define the probability distribution of the amount of time until some specific event takes place.

A random variable X follows an exponential distribution with parameter m.

The decay parameter is, m.

The probability distribution function of an Exponential distribution is:

$f(x)=me^{-mx}\ ;\ m>0, x>0$

Given: The decay parameter is, $\frac{1}{20}$

X is defined as the distance people are willing to commute in miles.

The decay parameter is m = $\frac{1}{20}$ .
The mean of the distribution is: $\mu=\frac{1}{m}=\frac{1}{\frac{1}{20}}=20$ .
The standard deviation is: $\sigma=\sqrt{variance}= \sqrt{\frac{1}{(m)^{2}} } =\frac{1}{m} =\frac{1}{\frac{1}{20}} =20$

Compute the probability that a person is willing to commute more than 25 miles as follows:

$P(X>25)=\int\limits^{\infty}_{25} {\frac{1}{20} e^{-\frac{1}{20}x}} \, dx \\=\frac{1}{20}|20e^{-\frac{1}{20}x}|^{\infty}_{25}\\=|e^{-\frac{1}{20}x}|^{\infty}_{25}\\=e^{-\frac{1}{20}\times25}\\=0.2865$