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3 years ago

Please Help!!! Determine the range of the following graph:

Mathematics

1 answer:

ELEN [110]3 years ago

5 0

Answer:

[-1, 6)

General Formulas and Concepts:

<u>Algebra I</u>

Reading a coordinate plane

<u>Algebra II</u>

Range is the set of y-values that are outputted by function f(x)
Interval Notation - [Brackets] denote inclusion and (Parenthesis) denote exclusion

Step-by-step explanation:

According to the graph, our y-values span from -1 to 6. Since -1 is a closed dot, it is inclusive. Since 6 is an open dot, it is exclusive:

Range would then be [-1, 6).

You might be interested in

Write an expression to represent the length of a string in yards in terms of the string's length, x, in inches

Sergio039 [100]

Answer:

$y = \frac{x}{36}$

Step-by-step explanation:

Given

Length of string = x inches

Required

Determine the length in yards

Represent the equivalent length with y

Such that

$y = x\ inches$

Divide x by 36 to get equivalent in yards

$y = \frac{1}{36} * x$

$y = \frac{x}{36}$

Hence, the equivalent in yards is $\frac{x}{36}$

3 0

3 years ago

If f(x) = 2x+7 and g(x) = -3x+5, what is f(g(1))?

Murrr4er [49]

Answer:

$f(g(1))=11$

Step-by-step explanation:

So we have the two functions:

$f(x)=2x+7\text{ and } g(x)=-3x+5$

And we want to find f(g(1)).

So, let's find g(1) first:

$g(x)=-3x+5$

Substitute 1 for x:

$g(1)=-3(1)+5$

Simplify:

$g(1)=-3+5$

Add:

$g(1)=2$

So:

$f(g(1))=f(2)$

Now, substitute 2 for x in f(x):

$f(x)=2x+7\\f(2)=2(2)+7$

Multiply:

$f(2)=4+7$

Add:

$f(2)=11$

So:

$f(g(1))=11$

8 0

3 years ago

50 points! I will mark brainliest! Random answers will be reported :)

ASHA 777 [7]

Answer:

$x=10$

Step-by-step explanation:

Notice that all three sides of the given triangle has one tick mark.

This means that all three sides are congruent, meaning that we have an equilateral triangle.

And for equilateral triangles, all three angles are also congruent.

So:

$\angle CBA=\angle BAC=\angle ACB$

Also, recall that the interior angles of a triangle is always 180. Thus:

$\angle CBA+\angle BAC+\angle ACB=180$

Substitute:

$\angle CBA+\angle CBA+\angle CBA=180$

Combine like terms:

$3\angle CBA=180$

Substitute:

$3(6x)=180$

Multiply:

$18x=180$

Thus:

$x=10$

And we are done!

3 0

3 years ago

There are 15 students in the 8th grade. The students are randomly placed into three different algebra classes of 5 students each

xeze [42]

Answer:

The probability is $\frac{6}{91}$ ≅ $0.0659$

Step-by-step explanation:

Let's analyze the question.

There are 15 students in the 8th grade.

The students are randomly placed into three different algebra classes of 5 students each.

We are looking for the probability that Trevor, Terry and Evan will be in the same algebra class.

One possible way to solve this question is to think about the product probability rule.

We can use it because we are in an equiprobable space. (And also the events are independent).

Let's set for example a class for Evan.

The probability that Evan will be in a class is $1$

Then for Terry there are $4$ places out of $14$ that puts Terry in the Evan's class.

We write $1.\frac{4}{14}$

Finally for Trevor there are $3$ places out of the remaining $13$ that puts Trevor in the same class with Evan and Terry.

Using the product rule we write :

$1.\frac{4}{14}.\frac{3}{13}=\frac{6}{91}$

The probability of the event is $\frac{6}{91}$ ≅ $0.0659$

5 0

3 years ago

A major department store chain is interested in estimating the mean amount its credit card customers spent on their first visit

sergiy2304 [10]

Answer:

95% Confidence interval: (39.43, 61.58)

Step-by-step explanation:

We are given the following in the question:

Sample mean, $\bar{x}$ = $50.50

Sample size, n = 15

Alpha, α = 0.05

Sample standard deviation = 20

95% Confidence interval:

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$

Putting the values, we get,

$t_{critical}\text{ at degree of freedom 14 and}~\alpha_{0.05} = \pm 2.1447$

$=50.50 \pm 2.1447(\dfrac{20}{\sqrt{15}} ) \\\\= 50.50 \pm 11.0751 \\= (39.4249,61.5751)\\\approx (39.43, 61.58)$

95% Confidence interval: (39.43, 61.58)

8 0

3 years ago