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dlinn [17]
3 years ago
9

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).

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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
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