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Rudik
2 months ago
10

What is the solution to –4(8 – 3x) ≥ 6x – 8?

Mathematics
2 answers:
AnnZ [12.3K]2 months ago
6 0

Response:

x ≥ 4

Step-by-step breakdown:

4(8 - 3x) ≥ 6x - 8  distribute the term in parentheses on the left side

32 + 12x ≥ 6x - 8 (subtract 6x from both sides)

32 + 6x ≥ - 8 (add 32 to both sides)

6x ≥ 24 (divide both sides by 6)

Thus, = x ≥ 4

tester [12.3K]2 months ago
5 0

Response:

x ≥ 4

Step-by-step breakdown:

Given

- 4(8 - 3x) ≥ 6x - 8 ← distribute the term in parentheses on the left side

- 32 + 12x ≥ 6x - 8 (subtract 6x from both sides)

- 32 + 6x ≥ - 8 (add 32 to both sides)

6x ≥ 24 (divide both sides by 6)

Thus, x ≥ 4

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He takes 6 hours to stitch a single shirt and stitches 6/10 of a shirt in 1 hour.
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L'età media di Aldo, Bruno, Carlo e Davide è 16 anni. Se non si tiene conto di Davide, l'età media dei tre rimanenti sale a 18.
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Answer:

Davide is 10 years old.

L'età di Davide è 10.

Step-by-step explanation:

The average ages of Aldo, Bruno, Carlo, and Davide amount to 16 years.

Let’s denote:

x for Aldo's age.

y for Bruno's age.

z for Carlo's age.

w for Davide's age.

The average for all four is 16 years.

This gives us:

16 = \frac{x + y + z + w}{4}

x + y + z + w = 64

Excluding Davide, the average age of the other three is 18. Therefore:

18 = \frac{x + y + z}{3}

x + y + z = 54

Substituting into the original equation:

x + y + z + w = 64

54 + w = 64

w = 10

Hence, Davide’s age is indeed 10.

L'età di Davide è 10.

4 0
2 months ago
Isabella built a time travel machine, but she can't control the destination of her trip. Each time she uses the machine she has
babunello [11817]

The likelihood that at least one trip occurs before Isabella's birth is 0.7627.

Step-by-step explanation:

In this scenario, Isabella has invented a time machine, but she lacks control over where she travels. Each use of the device holds a 0.25 probability of leading her to a time preceding her birth. Over the initial year of trials, she operates her machine 5 times. If we assume every journey has an equal chance of going back in time, we can calculate the odds that at least one of these trips occurs before she was born. Here's the calculation:

The probability of traveling to a time prior to her birth is 0.25.

The chance of not traveling back in time, given that the machine is used 5 times:

⇒ (1-0.25)(1-0.25)(1-0.25)(1-0.25)(1-0.25)

⇒ (1-0.25)^5

⇒ (0.75)^5

The probability that at least one trip goes before Isabella's birth is equal to 1 minus the probability of not traveling back to that period:

⇒ 1-(0.75)^5

⇒ 0.7627

Consequently, the chance that at least one trip travels before Isabella's birth is 0.7627.

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