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2 years ago

Identify the range of the function shown in the graph.

Mathematics

1 answer:

MAVERICK [17]2 years ago

3 0

$\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the Concept of the plots and graphs.

Since, the x axis varies from -infinite to +infinite,

here, the y totally varies from -5 to +5,

hence, the option B.) is correct.

===>

$- 5 \leqslant y \leqslant 5$

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Pleas help me out here.

Amiraneli [1.4K]

Answer:

A and E

Step-by-step explanation:

The given problem shows a graph of a vertical line with an undefined slope, in which its equation is x = 3 (which matches Option A).

It is not considered a function because it has the same input for every given output. This means that regardless of its y-coordinate, its corresponding x-coordinate will always be x = 3.

The Vertical Line Test allows us to know whether or not a graph is actually a function. Remember that a function can only take on one output for each input. We cannot plug in a value and produce two output values. Since the graph represents a vertical line, it automatically fails the vertical line test. Therefore, it is not a function.

Because it is not a function, Options B, C, and D are invalid answers.

Hence, the correct answers are Options A and E.

4 0

2 years ago

How to find upper and lower bound of a definite integral given an unknown equation?

worty [1.4K]

Finding the upper and lower bounds for a definite integral without an equation is pretty hard because how can we find the upper and lower bounds of definite integral if there is no equation given. But I will teach you how to find the lower and upper bounds of a definite integral, when the equation is like this

$\int\limits^6_1 t^{2} - 6t + 11dt$

So, i integrate this,
$( \frac{t^{3} }{3} - 3t^{2} + 11t) \int\limits^6_1$

I know I have a minimum at x=3 because;
f(t )= t^2 − 6t + 11
f′(t) = 2
t−6 = 0
2(t−3) = 0
t = 3
f(5) = 4
f(1) = −4

4 0

2 years ago

Plz help me with this!!!!

julia-pushkina [17]

Answer:

by the distance formula

the points are (2,2) and (-2,7)

and subtituting d=sqrt((2-(-2))^2+(2-7)^2)

which is equal to sqrt of 41

and it is equal to 6.40

3 0

2 years ago

Can someone please tell me what the equation for this will be.

Papessa [141]

Answer:

y = 0.8x + 0.5

Step-by-step explanation:

First, find the slope using two pairs of values, (0.5, 0.9) and (0.75, 1.1):

$slope (m) = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1.1 - 0.9}{0.75 - 0.5} = \frac{0.2}{0.25} = 0.8$

m = 0.8

Find y-intercept, b, by substituting x = 0.5, y = 0.9 and m = 0.8 into y = mx + b

Thus:

0.9 = (0.8)(0.5) + b

0.9 = 0.4 + b

0.9 - 0.4 = b

b = 0.5

To write the equation substitute m = 0.8, and b = 0.5 into y = mx + b.

Thus:

y = 0.8x + 0.5

7 0

2 years ago

Crime and Punishment: In a study of pleas and prison sentences, it is found that 45% of the subjects studied were sent to prison

Gala2k [10]

Answer:

(a) The probability of getting someone who was not sent to prison is 0.55.

(b) If a study subject is randomly selected and it is then found that the subject entered a guilty plea, the probability that this person was not sent to prison is 0.63.

Step-by-step explanation:

We are given that in a study of pleas and prison sentences, it is found that 45% of the subjects studied were sent to prison. Among those sent to prison, 40% chose to plead guilty. Among those not sent to prison, 55% chose to plead guilty.

Let the probability that subjects studied were sent to prison = P(A) = 0.45

Let G = event that subject chose to plead guilty

So, the probability that the subjects chose to plead guilty given that they were sent to prison = P(G/A) = 0.40

and the probability that the subjects chose to plead guilty given that they were not sent to prison = P(G/A') = 0.55

(a) The probability of getting someone who was not sent to prison = 1 - Probability of getting someone who was sent to prison

P(A') = 1 - P(A)

= 1 - 0.45 = 0.55

(b) If a study subject is randomly selected and it is then found that the subject entered a guilty plea, the probability that this person was not sent to prison is given by = P(A'/G)

We will use Bayes' Theorem here to calculate the above probability;

P(A'/G) = $\frac{P(A') \times P(G/A')}{P(A') \times P(G/A') +P(A) \times P(G/A)}$

= $\frac{0.55 \times 0.55}{0.55\times 0.55 +0.45 \times 0.40}$

= $\frac{0.3025}{0.4825}$

= <u>0.63</u>

6 0

2 years ago