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3 years ago

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In a classroom,there are 12 girls and 8 boys.What fraction of the students in class are girls?What percent of the class are girl

s?

1 answer:

tekilochka [14]3 years ago

4 0

Answer:

Step-by-step explanation:

Add the total number of students:

Boys: 8, + Girls: 12 = 20 students

$\frac{12}{20} = \frac{6}{10}$ of the students are girls

6/10 also equates to 60%, which is the percent of girls in the class.

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Pls help me with my math

givi [52]

Answer:

The definition for the given piecewise-defined function is: $\boxed{\displaystyle\sf\ Option\:D:\:\: f(x) = \begin{cases}\displaystyle\sf\ x + 2 & \sf\:{if\:\:x \leq -1} \\\displaystyle\sf\ 2x + 4 & \sf\:{if\:\:x > -1}\end{cases}}$ .

Step-by-step explanation:

<h3>General Concepts:</h3>

Piecewise-defined functions.
Interval notations.

<h3>What is a piecewise-defined function?</h3>

A piecewise-defined function represents specific rules over different intervals of the domain.

<h3>Symbols used in expressing interval notations:</h3>

Open interval: This means that the endpoint is <em>not</em> included in the interval.

We can use the following symbols to indicate the <u>exclusion</u> of endpoints in the interval:

Left or right parenthesis, "( )" (or both).

Greater than (>) or less than (<) symbols.
Open dot " $\circ$ " is another way of expressing the exclusion of an endpoint in the graph of a piecewise-defined function.

Closed interval: This implies the inclusion of endpoints in the interval.

We can use the following symbols to indicate the <u>inclusion</u> of endpoints in the interval:

Open- or closed brackets (or both), "[ ]."

Greater than or equal to (≥) or less than or equal to (≤) symbols.
Closed circle or dot, "•" is another way of expressing the <em>inclusion</em> of the endpoint in the graph of a piecewise-defined function.

<h2>Determine the appropriate function rule that defines different parts of the domain. </h2>

The best way to determine which piecewise-defined function represents the graph is by observing the <u>endpoints</u> and <u>orientation</u> of both partial lines.

Open circle on (-1, 2): The graph shows that one of the partial lines has an <em>excluded</em> endpoint of (-1, 2) extending towards the <u>right</u>. This implies that its domain values are defined when x > -1.

Closed circle on (-1, 1): The graph shows that one of the partial lines has an <em>included</em> endpoint of (-1, 1) extended towards the <u>left</u>. Hence, its domain values are defined when x ≤ -1.

Based on our observations from the previous step, we can infer that x > -1 or x ≤ -1 apply to piecewise-defined functions A or D. However, only one of those two options represent the graph.

<h2>Solution:</h2><h3>a) Test option A:</h3>

$\boxed{\displaystyle\sf Option\:A)\:\:\:f(x) = \begin{cases}\displaystyle\sf\ 2x + 2 & \sf\:{if\:\:x \leq -1} \\\displaystyle\sf\ x + 4 & \sf\:{if\:\:x > -1}\end{cases}}$

<h3>Piece 1: If x ≤ -1, then it is defined by f(x) = 2x + 2. </h3>

We must choose a domain value that falls within the interval of x ≤ -1 whose output is included is included in the graph of the partial line with a <u>closed dot</u>.

Substitute x = -2 into f(x) = 2x + 2:

f(x) = 2x + 2

f(-2) = 2(-2) + 2
f(-2) = -4 + 2
f(-2) = -2 ⇒ <em>False statement</em>.

⇒ The output value of f(-2) = -2 is <u>not</u> included in the graph of the partial line whose endpoint is at (-1, 1).

<h3>Piece 2: If x > -1, then it is defined by f(x) = x + 4. </h3>

We must choose a domain value that falls within the interval of x > -1 whose output is included in the graph of the partial line with an <u>open dot</u>.

Substitute x = 0 into f(x) = x + 4:

f(x) = x + 4
f(0) = (0) + 4
f(0) = 4 ⇒ <em>True statement</em>.

⇒ The output value of f(0) = 4 <u>is</u> included in the graph of the partial line whose endpoint is at (-1, 2).

Conclusion for Option A:

Option A is not the correct piecewise-defined function because one of the pieces, f(x) = 2x + 2, does not specify the interval (-∞, -1].

<h3>b) Test option D:</h3>

$\boxed{\displaystyle\sf Option\:D)\:\:\:f(x) = \begin{cases}\displaystyle\sf\ x + 2 & \sf\:{if\:\:x \leq -1} \\\displaystyle\sf\ 2x + 4 & \sf\:{if\:\:x > -1}\end{cases}}$

<h3>Piece 1: If x ≤ -1, then it is defined by f(x) = x + 2. </h3>

We must choose a domain value that falls within the interval of x ≤ -1 whose output is included is included in the graph of the partial line with a <u>closed dot</u>.

Substitute x = -2 into f(x) = x + 2:

f(x) = x + 2
f(-2) = (-2) + 2
f(-2) = 0 ⇒ <em>True statement</em>.

⇒ The output value of f(-2) = 0 <u>is</u> included the graph of the partial line whose endpoint is at (-1, 1).

<h3>Piece 2: If x > -1, then it is defined by f(x) = 2x + 4.</h3>

We must choose a domain value that falls within the interval of x > -1 whose output is included is included in the graph of the partial line with an <u>open dot</u>.

Substitute x = 0 into f(x) = 2x + 4:

f(x) = 2x + 4
f(0) = 2(0) + 4
f(0) = 0 + 4 = 0 ⇒ <em>True statement</em>.

⇒ The output value of f(0) = 4 <u>is</u> included in the graph of the partial line whose endpoint is at (-1, 2).

<h2>Final Answer: </h2>

We can infer that the piecewise-defined function that represents the graph is:

$\boxed{\displaystyle\sf\ Option\:D:\:\: f(x) = \begin{cases}\displaystyle\sf\ x + 2 & \sf\:{if\:\:x \leq -1} \\\displaystyle\sf\ 2x + 4 & \sf\:{if\:\:x > -1}\end{cases}}$ .

________________________________________

Learn more about piecewise-defined functions here:

brainly.com/question/26145479

8 0

2 years ago

Determine,in each of the following cases, whether the described system is or not a group. Explain your answers. Determine what i

zheka24 [161]

Answer:

(a) Not a group

(b) Not a group

(c) Abelian group

Step-by-step explanation:

<em>In order for a system <G,*> to be a group, the following must be satisfied </em>

<em> (1) The binary operation is associative, i.e., (a*b)*c = a*(b*c) for all a,b,c in G </em>

<em>(2) There is an identity element, i.e., there is an element e such that a*e = e*a = a for all a in G </em>

<em> (3) For each a in G, there is an inverse, i.e, another element a' in G such that a*a' = a'*a = e (the identity) </em>

<em> </em>

If in addition the operation * is commutative (a*b = b*a for every a,b in G), then the group is said to be Abelian

(a)

The system <G,*> is not a group since there are no identity.

To see this, suppose there is an element e such that

a*e = a

then

a-e = a which implies e=0

It is easy to see that 0 cannot be an identity.

For example

2*0 = 2-0 = 2

Whereas

0*2 = 0-2 = -2

So 2*0 is not equal to 0*2

(b)

The system <G,*> is not a group either.

If A is a matrix 2x2 and the determinant of A det(A)=0, then the inverse of A does not exist.

(c)

The table of the operation G is showed in the attachment.

It is evident that this system is isomorphic under the identity map, to the cyclic group

$\mathbb{Z}_{5}$

the system formed by the subset of Z, {0,1,2,3,4} with the operation of addition module 5, which is an Abelian cyclic group

We conclude that the system <G,*> is Abelian.

Attachment: Table for the operation * in (c)

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3 years ago

(-2) • (-4.5) <br> answer

sashaice [31]

Answer:

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plz mark brainliest

Step-by-step explanation:

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3 years ago

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On the first day of a family road trip, Tamara's family travels 540 miles,

konstantin123 [22]

Answer:

the right answer is 486 and if I am right can brainlist

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2 years ago

What are “like terms”? Why can we only add like terms?

SSSSS [86.1K]

Answer:

Like terms are terms whose variables are the same.

We can only add them because if they does not have the same variable you do not actually know if the variables are equivalent and you will get the wrong answer.

Step-by-step explanation:

Ex: 4z,7z,108z,45z

Ex: 4w-2h+7a-2w

You can not add 4w and 7a because you do not know what the variables equal so you do not know the true number so add or subtract by.

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3 years ago