All categories

1 year ago

The graph of a function h is shown below.

Mathematics

1 answer:

kipiarov [429]1 year ago

3 0

According to the graph of the function, it is found that:

a) h(2) = 1.

b) h(x) = -3 when x = -4.

Looking at the graph, h(2) is given by the value of y when x = 2, that is, the value of the vertical axis when the horizontal axis is of 2. Hence, h(2) = 1.

<h3>What is the value of x for which h(x) = -3?</h3>

Looking at the graph, it is given by x when y = -3, that is, the value of the horizontal axis when the vertical axis is of -3, hence, h(x) = -3 when x = -4.

More can be learned about functions at brainly.com/question/25537936

You might be interested in

Calculate:|a|−|b|+|c|, if a=−8; b=−5; c=1

Fynjy0 [20]

Answer:

The answer is 4.

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Help<br><br> The diameter of the circle is 8in

-BARSIC- [3]

I believe your answer is 25.12

7 0

2 years ago

Read 2 more answers

What value of x makes this equation true?<br> 0.5x – 8= 0.2x + 7

postnew [5]

The value of x that will make this equation true is 50.

0.5 times 50=25 -8 = 17

0.2 times 50= 10 +7 = 17

Therefore, your answer is 50.

3 0

3 years ago

Verify that the following function is a probability mass function, and determine the requested probabilities. f left-parenthesis

Licemer1 [7]

Answer:

$f(x) = \frac{27}{13} (\frac{1}{3})^x , x=1,2,3$

1) $f(x_i) \geq 0, \forall x_i$

2) $sum_{i=1}^n P(X_i) =1$

We can find the individual probabilities and we got:

$f(1)= \frac{27}{13} (\frac{1}{3})^1 =0.6923$

$f(2)= \frac{27}{13} (\frac{1}{3})^2 =0.2307$

$f(3)= \frac{27}{13} (\frac{1}{3})^3 =0.0770$

And the sum of the 3 values 0.6923+0.2307+0.0770= 1 so then we satisfy all the conditions and we can conclude that f(x) is a probability distribution.

$P(X \leq 1) = P(X=1) =0.6923$

$P(X>1) = P(X=2) +P(X=3) = 0.2307+0.0770=0.3077$

$P(2$

Step-by-step explanation:

For this case we have the following density function:

$f(x) = \frac{27}{13} (\frac{1}{3})^x , x=1,2,3$

In order to satisfty that this function is a probability mass function we need to check two conditions:

1) $f(x_i) \geq 0, \forall x_i$

2) $sum_{i=1}^n P(X_i) =1$

We can find the individual probabilities and we got:

$f(1)= \frac{27}{13} (\frac{1}{3})^1 =0.6923$

$f(2)= \frac{27}{13} (\frac{1}{3})^2 =0.2307$

$f(3)= \frac{27}{13} (\frac{1}{3})^3 =0.0770$

And the sum of the 3 values 0.6923+0.2307+0.0770= 1 so then we satisfy all the conditions and we can conclude that f(x) is a probability distribution.

And if we want to find the following probabilities:

$P(X \leq 1) = P(X=1) =0.6923$

$P(X>1) = P(X=2) +P(X=3) = 0.2307+0.0770=0.3077$

$P(2$

7 0

2 years ago

Domain dan range dari f(x)=√3x-6

hodyreva [135]

Domain: [2,∞)
Range: [0, ∞)

hope it helps!

4 0

2 years ago