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3 years ago

Solve for y 6x-8y>=-24

1 answer:

3 0

$6x-8y\geq-24\qquad\text{subtract 6x from both sides}\\\\-8y\geq-6x-24\qquad\text{change the signs}\\\\8y\geq6x+24\qquad\text{divide both sides by 8}\\\\y\geq\dfrac{6}{8}x+3\\\\\boxed{y\geq\dfrac{3}{4}x+3}$

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A manufacturer plans on building a better mousetrap. There is a 60% chance the mouse trap will produce a profit of $200,000 for

maksim [4K]

Expected value E(x) = 0.6(200,000) + 0.2(100,000) - 0.2(200,000) = 120,000 + 20,000 - 40,000 = 100,000

Therefore, the expected value is $100,000, so the company should build the mousetrap.

4 0

3 years ago

Read 2 more answers

Do the product 40x500 and 40x600 have the same number of zeros

VARVARA [1.3K]

No they do not have the same number of zeroes

Although the 2 products do seem like they might have the same number of zeroes, 40 x 500 and 40 x 600

both have 3 zeroes before multiplying

if we take the first 2 digits of the 2 numbers being multiplied to get an idea of the answer

if we take 40 x 500 - 4 x 5 = 20

then theres an extra zero added

but if we take 40 x 600 - 4 x 6 = 24

no extra zeroes added to this

so if we finally look at the answers

40 x 500 = 20 000

40 x 600 = 24 000

20000 has 4 zeroes and 24000 has 3 zeroes

so the answer is no they dont have the same number of zeroes

8 0

3 years ago

Read 2 more answers

How do I find the area of a rhombus?

QveST [7]

A = pq/2
p = diagonal 1

q = diagonal 2

5 0

3 years ago

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The short sides of a rectangle are 2 inches. The long sides of the same rectangle are three less than an unknown number of inche

Amiraneli [1.4K]

Answer

The first student was right.

The length of the long side is at least 13 inches.

Explanation

The perimeter of any figure is the distance all round.

Perimeter of a rectangle = 2(l+w). Where l is length and w is the width.

2(l + w) ≥ 30

2{(x-3) + 2} ≥ 30

2(x-3+2) = 30

2(x - 1) ≥ 30

x - 1 ≥ 15

x ≥ 16

When x = 16,

l = 16-3

= 13 inches

The length of the long side is 13 inches. The first student was right.

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3 years ago

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In a certain Algebra 2 class of 30 students, 19 of them play basketball and 12 of them play baseball. There are 8 students who p

Alenkinab [10]

Answer:

Probability that a student chosen randomly from the class plays basketball or baseball is $\frac{23}{30}$ or 0.76

Step-by-step explanation:

Given:

Total number of students in the class = 30

Number of students who plays basket ball = 19

Number of students who plays base ball = 12

Number of students who plays base both the games = 8

To find:

Probability that a student chosen randomly from the class plays basketball or baseball=?

Solution:

$P(A \cup B)=P(A)+P(B)-P(A \cap B)$ ---------------(1)

where

P(A) = Probability of choosing a student playing basket ball

P(B) = Probability of choosing a student playing base ball

P(A \cap B) = Probability of choosing a student playing both the games

Finding P(A)

P(A) = $\frac{\text { Number of students playing basket ball }}{\text{Total number of students}}$

P(A) = $\frac{19}{30}$ --------------------------(2)

Finding P(B)

P(B) = $\frac{\text { Number of students playing baseball }}{\text{Total number of students}}$

P(B) = $\frac{12}{30}$ ---------------------------(3)

Finding $P(A \cap B)$

P(A) = $\frac{\text { Number of students playing both games }}{\text{Total number of students}}$

P(A) = $\frac{8}{30}$ -----------------------------(4)

Now substituting (2), (3) , (4) in (1), we get

$P(A \cup B)= \frac{19}{30} + \frac{12}{30} -\frac{8}{30}$

$P(A \cup B)= \frac{31}{30} -\frac{8}{30}$

$P(A \cup B)= \frac{23}{30}$

7 0

3 years ago