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3 years ago

Find the solution set to each inequality. Express the solution in set notation.

Mathematics

1 answer:

daser333 [38]3 years ago

3 0

The solution to the inequality 6m + 2 > -27 is m > -4.33

The solution to the inequality 8(p-6)>4(p-4) is p > 8

The given inequality is:

6m + 2 > - 27

Subtract 2 to both sides of the inequality

6m + 2 - 2 > -27 - 2

6m > -29

Divide both sides by 6

$\frac{6m}{6}> \frac{-29}{6} \\m > \frac{-29}{6}\\m > -4.83$

For the inequality 8(p-6)>4(p-4)

Expand the inequality using the distributive rule

8p - 48 > 4p - 16

Collect like terms

8p - 4p > -16 + 48

4p > 32

Divide both sides of the inequality 4

$\frac{4p}{4} > \frac{32}{4}\\p > 8$

The solution to the inequality 6m + 2 > -27 is m > -4.33

The solution to the inequality 8(p-6)>4(p-4) is p > 8

Learn more here: brainly.com/question/15816805

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A packing box is being filled with packing peanuts. The box measures 1 1/2 feet by 1 1/2 feet by 2 feet. How many cubic feet of

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Answer: $4.5\ ft^3$

Step-by-step explanation:

Given

Measurement of box $1\frac{1}{2}\times 1\frac{1}{2}\times 2$

suppose Length $L=1\frac{1}{2}=\frac{3}{2}\ ft$

Breadth $B=1\frac{1}{2}=\frac{3}{2}\ ft$

height $H=2\ ft$

Volume of Cuboid $=L\times B\times H$

Volume of Packaging box $V=\frac{3}{2}\times \frac{3}{2}\times 2$

$V=\frac{9}{2}\ ft^3$

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4 years ago

A bag contains 4 blue marbles and 2 yellow marbles. Two marbles are randomly chosen (the first marble is NOT replaced before dra

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Answer:

0.4 = 40% probability that both marbles are blue.

0.0667 = 6.67% probability that both marbles are yellow.

53.33% probability of one blue and then one yellow

If you are told that both selected marbles are the same color, 0.8571 = 85.71% probability that both are blue

Step-by-step explanation:

To solve this question, we need to understand conditional probability(for the final question) and the hypergeometric distribution(for the first three, because the balls are chosen without being replaced).

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

What is the probability that both marbles are blue?

4 + 2 = 6 total marbles, which means that $N = 6$

4 blue, which means that $k = 4$

Sample of 2, which means that $n = 2$

This is P(X = 2). So

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 2) = h(2,6,2,4) = \frac{C_{4,2}*C_{2,0}}{C_{6,2}} = 0.4$

0.4 = 40% probability that both marbles are blue

What is the probability that both marbles are yellow?

4 + 2 = 6 total marbles, which means that $N = 6$

2 yellow, which means that $k = 2$

Sample of 2, which means that $n = 2$

This is P(X = 2). So

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 2) = h(2,6,2,2) = \frac{C_{2,2}*C_{4,0}}{C_{6,2}} = 0.0667$

0.0667 = 6.67% probability that both marbles are yellow.

What is the probability of one blue and then one yellow?

Total is 100%.

Can be:

Both blue(40%)

Both yellow(6.67%)

One blue and one yellow(x%). So

40 + 6.67 + x = 100

x = 100 - 46.67

x = 53.33

53.33% probability of one blue and then one yellow.

If you are told that both selected marbles are the same color, what is the probability that both are blue?

Conditional probability.

Event A: Both same color

Event B: Both blue

Probability of both being same color:

Both blue(40%)

Both yellow(6.67%)

This means that $P(A) = 0.4 + 0.0667 = 0.4667$

Probability of both being the same color and blue:

40% probability that both are blue, which means that $P(A \cap B) = 0.4$

Desired probability:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.4}{0.4667} = 0.8571$

If you are told that both selected marbles are the same color, 0.8571 = 85.71% probability that both are blue

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The product would expand to

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This is a trinomial, and the only way to make it a binomial is to cancel out a coefficient using our variable $x$ .

So, we can cancel either the linear term or the constant term.

In the first case, we require

$7x-50=0 \iff 7x=50 \iff x=\dfrac{50}{7}=\dfrac{49}{7}+\dfrac{1}{7}=7\dfrac{1}{7}$

In the second case, we require

$-10x=0\iff 10x=0 \iff x=0$

But $x$ must be a non-zero rational number, so this solution is not feasible.

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