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Mekhanik [1.2K]

3 years ago

12

If 9\geq4x+19≥4x+19, is greater than or equal to, 4, x, plus, 1, which inequality represents the possible range of values of 12x

+312x+312, x, plus, 3?

1 answer:

AleksAgata [21]3 years ago

6 0

Answer:

$12x+3\leq 27$

Step-by-step explanation:

It is given that

$9\geq4x+1$

We need to find the possible range of values of 12x+3.

Subtract -1 from both sides.

$9-1\geq 4x$

$8\geq 4x$

Multiply both sides by 3.

$24\geq 12x$

Add 3 on both sides.

$24+3\geq 12x+3$

$27\geq 12x+3$

It can be rewritten as

$12x+3\leq 27$

Therefore, 12x+3 must be less than or equal to 27.

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Which expression is equivalent to 3m+1-m?

Lisa [10]

Answer:

option (a) is correct.

2 + m - 1 + m is an equivalent expression to the given expression 3m+1-m

Step-by-step explanation:

Given expression 3m+1-m

We have to choose an equivalent expression from given options.

Equivalent expression are those expression that looks different but are same.

Like 4+2 = 6 and 3+ 3 = 6

Both have same value but looks differently.

Like terms are term having same variable with same degree.

Consider the given expression 3m+1-m

Simplify by adding like terms,

3m + 1 - m = (3-1) m + 1

Thus, (3-1) m + 1 = 2m + 1

Also 2m + 1 can be written as m + m + 2 - 1

Thus, option (a) is correct.

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A movie theater charges $15 for standard viewing, $20 for 3D viewing, and $35 for Dinner and a

Ludmilka [50]

The quantity of each type of seats sold are as follows:

Standard viewing = 2000

3D seats = 800

Movie and a dinner seat = 200

According to the question,there are four times as many 3D, x seats as Dinner and a Movie, y seats.

That is, x = 4y

Also, total seats

= (x) + (y) + (z) = 3000...….............eqn(1)

Also, If the theater brings in $53,000 when tickets to all 3000 seats are sold.

20x + 35y + 15z = 53000...........eqn(2)

By substituting 4y for x in equations 1 and 2; we have;

<em>5y + z = 3000</em>..…........eqn(3) and

<em>115y + 15z = 53000</em>.........eqn(4)

By solving equations 3 and 4 simultaneously; we have;

y = 200 and z = 2000

and since x = 4y

x = 800

The quantity of each type of seats sold is as follows:

Standard viewing = 2000

3D seats = 800

Movie and a dinner seat = 200

Read more:

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What is the square root of 96<br><br><br><br><br> Follow my new brainly at: fol

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Answer:

4√6

Step-by-step explanation:

For right now, I will assume that the "<em>Follow my new brainly at..</em>" part is irrelevant.

Anyways, to find the square root of a number, usually the best way is to find its prime factorization first:

96 : $2^5\cdot \:3$

We can rewrite that as:

$\sqrt{2^4\cdot \:2\cdot \:3}$

Simplify!

$2^{\frac{4}{2}}\sqrt{2\cdot \:3}$

Since $2^{\frac{4}{2}}\sqrt{2\cdot \:3}$ is the same thing as $4\sqrt{6}$ , that is our answer.

Thusly, as seen above, the answer is $\boxed{4\sqrt{6}}$ .

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$4.49 divided by 16 in PLS help

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It would be $0.28 for each inch.

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3 years ago

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Three couples and two single individuals have been invited to an investment seminar and have agreed to attend. Suppose the proba

Maksim231197 [3]

Answer:

(a) Probability mass function

P(X=0) = 0.0602

P(X=1) = 0.0908

P(X=2) = 0.1704

P(X=3) = 0.2055

P(X=4) = 0.1285

P(X=5) = 0.1550

P(X=6) = 0.1427

P(X=7) = 0.0390

P(X=8) = 0.0147

NOTE: the sum of the probabilities gives 1.0068 for rounding errors. It can be divided by 1.0068 to get the adjusted values.

(b) Cumulative distribution function of X

F(X=0) = 0.0602

F(X=1) = 0.1510

F(X=2) = 0.3214

F(X=3) = 0.5269

F(X=4) = 0.6554

F(X=5) = 0.8104

F(X=6) = 0.9531

F(X=7) = 0.9921

F(X=8) = 1.0068

Step-by-step explanation:

Let X be the number of people who arrive late to the seminar, we can assess that X can take values from 0 (everybody on time) to 8 (everybody late).

<u>For X=0</u>

This happens when every couple and the singles are on time (ot).

$P(X=0)=P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot)\\\\P(X=0)=(1-0.43)^{5}=0.57^5= 0.0602$

<u>For X=1</u>

This happens when only one single arrives late. It can be #4 or #5. As the probabilities are the same (P(#4=late)=P(#5=late)), we can multiply by 2 the former probability:

$P(X=1) = P(\#4=late)+P(\#5=late)=2*P(\#4=late)\\\\P(X=1) = 2*P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=late)*P(\#5=ot)\\\\P(X=1) = 2*0.57*0.57*0.57*0.43*0.57\\\\P(X=1) = 2*0.57^4*0.43=2*0.0454=0.0908$

<u>For X=2</u>

This happens when

1) Only one of the three couples is late, and the others cooples and singles are on time.

2) When both singles are late , and the couples are on time.

$P(X=2)=3*(P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot))+P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=l)\\\\P(X=2)=3*(0.43*0.57^4)+(0.43^2*0.57^3)=0.1362+0.0342=0.1704$

<u>For X=3</u>

This happens when

1) Only one couple (3 posibilities) and one single are late (2 posibilities). This means there are 3*2=6 combinations of this.

$P(X=3)=6*(P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=3)=6*(0.43^2*0.57^3)=6*0.342=0.2055$

<u>For X=4</u>

This happens when

1) Only two couples are late. There are 3 combinations of these.

2) Only one couple and both singles are late. Only one combination of these situation.

$P(X=4)=3*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot))+P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=l)\\\\P(X=4)=3*(0.43^2*0.57^3)+(0.43^3*0.57^2)\\\\P(X=4)=3*0.0342+ 0.0258=0.1027+0.0258=0.1285$

<u>For X=5</u>

This happens when

1) Only two couples (3 combinations) and one single are late (2 combinations). There are 6 combinations.

$P(X=6)=6*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=6)=6*(0.43^3*0.57^2)=6*0.0258=0.1550$

<u>For X=6</u>

This happens when

1) Only the three couples are late (1 combination)

2) Only two couples (3 combinations) and one single (2 combinations) are late

$P(X=6)=P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=ot)*P(\#5=ot)+6*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=6)=(0.43^3*0.57^2)+6*(0.43^4*0.57)\\\\P(X=6)=0.0258+6*0.0195=0.0258+0.1169=0.1427$

<u>For X=7</u>

This happens when

1) Only one of the singles is on time (2 combinations)

$P(X=7)=2*P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=l)*P(\#5=ot)\\\\P(X=7)=2*0.43^4*0.57=0.0390$

<u>For X=8</u>

This happens when everybody is late

$P(X=8)=P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=l)*P(\#5=l)\\\\P(X=8) = 0.43^5=0.0147$

8 0

3 years ago