4 years ago

Identify the value that is NOT in range of the piecewise function. A. -3 B. -2 C. 3

Mathematics

1 answer:

Blizzard [7]4 years ago

7 0

Answer:

Option A i.e. y = - 3 is not in range of the piece wise function graphed.

Step-by-step explanation:

The range of a function y = f(x) is given by the possible values of y for the possible domain (x-values) of the function.

Now, it is clear from the graph of the function that the value of y never goes below y = - 2.

The graph is drawn in such a way that any value of y less than - 2 and greater than - ∞ is not in the range of the function.

Therefore, option A i.e. y = - 3 is not in range of the piecewise function graphed. (Answer)

You might be interested in

6 * 10^8 is how many times as large as 2 * 10^3

REY [17]

Answer:

300000 or 3 * 10^5

6 0

3 years ago

Read 2 more answers

There are 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 bacteria cells after 3 hours. Write the repeated multiplication as a power and then

sukhopar [10]

Answer:

2 over the power of 3

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

The scissor lift is a device that can lower and raise a platform. the scissor lift can raise the platform at a rate of 2.0 inche

Svetradugi [14.3K]

Answer:

i thank it is 25 seconds.

Step-by-step explanation:

3 0

4 years ago

29 points!!!Find the length of ac<br> Ab=36 cm express your answer in pi

FromTheMoon [43]

Mistake on my part ignore this

5 0

4 years ago

Read 2 more answers

The probability that your call to a service line is answered in less than 30 seconds is 0.75. Assume that your calls are indepen

e-lub [12.9K]

Answer with Step-by-step explanation:

Since the given event is binary we can use Bernoulli's probability to sove the problem

Thus for an event 'E' with probability of success 'p' the probability that the event occurs 'r' times in 'n' trails is given by

$P(E)=\frac{n!}{(n-r)!\cdot r!}\cdot p^{r}\cdot (1-p)^{n-r}$

Part a)

For part a n = 11 , r =9, p = 0.75

Applying values we get

$P(E)=\frac{11!}{(11-9)!\cdot 9!}\cdot (0.75)^{9}\cdot (1-0.75)^{11-9}\\\\\therefore P(E)=0.2581$

Part b)

For part b n = 20 , r = 16 , p=0.75

Applying values we get

$P(E)=\frac{20!}{(20-16)!\cdot 16!}\cdot (0.75)^{16}\cdot (1-0.75)^{20-16}\\\\\therefore P(E)=0.1896$

3 0

3 years ago