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konstantin123 [22]
3 years ago
12

Use the inequality 24 ≥ 58 + 5(x - 3.8)

Mathematics
1 answer:
spayn [35]3 years ago
4 0

Answer:

Work shown below!

Step-by-step explanation:

24 ≥ 58 + 5(x - 3.8)

Distribute;

24 ≥ 58 + 5x - 19

Collect like terms;

24 ≥ 39 + 5x

Subtract 39 from both sides;

-15 ≥ 5x

Divide both sides by 5;

x ≥ -3

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Answer:

50,000   fifty-thousand

Step-by-step explanation:

Just use a calculator

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2 years ago
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Firlakuza [10]

Answer:

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Step-by-step explanation: hope that answers your question :)

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kirk has 42 pieces of candy to divide evenly between his three children. if he puts the pieces into three boxes how many pieces
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3 years ago
In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A
vredina [299]

Answer:

c) P(270≤x≤280)=0.572

d) P(x=280)=0.091

Step-by-step explanation:

The population of bearings have a proportion p=0.90 of satisfactory thickness.

The shipments will be treated as random samples, of size n=500, taken out of the population of bearings.

As the sample size is big, we will model the amount of satisfactory bearings per shipment as a normally distributed variable (if the sample was small, a binomial distirbution would be more precise and appropiate).

The mean of this distribution will be:

\mu_s=np=500*0.90=450

The standard deviation will be:

\sigma_s=\sqrt{np(1-p)}=\sqrt{500*0.90*0.10}=\sqrt{45}=6.7

We can calculate the probability that a shipment is acceptable (at least 440 bearings meet the specification) calculating the z-score for X=440 and then the probability of this z-score:

z=(x-\mu_s)/\sigma_s=(440-450)/6.7=-10/6.7=-1.49\\\\P(z>-1.49)=0.932

Now, we have to create a new sampling distribution for the shipments. The size is n=300 and p=0.932.

The mean of this sampling distribution is:

\mu=np=300*0.932=279.6

The standard deviation will be:

\sigma=\sqrt{np(1-p)}=\sqrt{300*0.932*0.068}=\sqrt{19}=4.36

c) The probability that between 270 and 280 out of 300 shipments are acceptable can be calculated with the z-score and using the continuity factor, as this is modeled as a continuos variable:

P(270\leq x\leq280)=P(269.5

d) The probability that 280 out of 300 shipments are acceptable can be calculated using again the continuity factor correction:

P(X=280)=P(279.5

8 0
3 years ago
3. Which of the following quadratic equations has no solution? A) 0 = −2(x − 5)2 + 3 B) 0 = −2(x − 5)(x + 3) C) 0 = 2(x − 5)2 +
klemol [59]

Answer:

D) 0 = 2(x + 5)(x + 3)

Step-by-step explanation:

Which of the following quadratic equations has no solution?

We have to solve the Quadratic equation for all the options in other to get a positive value as a solution for x.

A) 0 = −2(x − 5)2 + 3

0 = -2(x - 5) × 5

0 = (-2x + 10) × 5

0 = -10x + 50

10x = 50

x = 50/10

x = 5

Option A has a solution of 5

B) 0 = −2(x − 5)(x + 3)

Take each of the factors and equate them to zero

-2 = 0

= 0

x - 5 = 0

x = 5

x + 3 = 0

x = -3

Option B has a solution by one of its factors as a positive value of 5

C) 0 = 2(x − 5)2 + 3

0 = 2(x - 5) × 5

0 = (2x -10) × 5

0 = 10x -50

-10x = -50

x = -50/-10

x = 5

Option C has a solution of 5

D) 0 = 2(x + 5)(x + 3)

Take each of the factors and equate to zero

0 = 2

= 0

x + 5 = 0

x = -5

x + 3 = 0

x = -3

For option D, all the values of x are 0, or negative values of -5 and -3.

Therefore the Quadratic Equation for option D has no solution.

3 0
3 years ago
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