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Elenna [48]
2 years ago
7

Which of the following is the solution to 3I x-1 I≥ 12?

Mathematics
1 answer:
hichkok12 [17]2 years ago
5 0

The solution for the given inequality will be x≤ -3 or x≥ 5. Thus, the correct option is B.

<h3>What are inequalities?</h3>

Inequalities help us to compare two unequal expressions. Also, it helps us to compare the non-equal expressions so that an equation can be formed. It is mostly denoted by the symbol <, >, ≤, and ≥.

For the given inequality the solution can be found as,

  • When the value of x is negative,

         3|x-1| ≥ 12

         x-1 ≥ 4

         x ≥ 5

  • When the value of x is positive,

         3|x-1| ≥ 12

         - x + 1 ≥ 4

         -x ≥ 3

         x ≤ -3

Hence, the solution for the given inequality will be x≤ -3 or x≥ 5. Thus, the correct option is B.

Learn more about Inequality:

brainly.com/question/19491153

#SPJ1

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Heeelllppp me please :)
Sophie [7]
The common ratio is 3.

I hope this helps!
4 0
3 years ago
suppose you have a dime, two pennies, and a quarter. one of the pennies was minted in 1976, and the other one was minted in 1992
notsponge [240]

The reason that the answers in part (a) and part (b) are not the same is because Q1 and Q2 are treated as two different coins in part (a) whereas they are effectively treated as the same coin in part (b). The reason they are treated as one coin is because they both contribute the same amount to the sum of money.

<h3>How to Solve Counting Problems?</h3>

A) If you choose at least one coin, this means you could choose 1, 2, 3 or 4 coins. Let us label the dime as D, the penny as P, the quarter minted in 1976 as Q1 and the quarter minted in 1992 as Q2.

Now, if you choose one coin, you could choose either D, P, Q1, or Q2. This gives us 4 possible sets.

If you choose two coins, you choose the following sets of coins: DP, DQ1, DQ2, PQ1, PQ2, Q1Q2. This gives us 6 possible sets.

If you choose three coins, you could the following sets of coins: DPQ1, DPQ2, DQ1Q2, PQ1Q2. This gives us 4 possible sets.

If you choose four coins, you can only choose DPQ1Q2. This gives us 1 possible set.

Therefore, the total number of different sets of coins you can form is 4 + 6 + 4 + 1 = 15 different sets of coins can be formed.

b) If you choose at least one coin, this means you could choose 1, 2, 3 or 4 coins.

If you choose one coin you could choose either D, P, Q1, or Q2. However, since Q1 and Q2 give us the same sum, they are effectively the same set. This gives us 3 possible sums (ten cents, one cent, or twenty-five cents.)

If you choose two coins, you could choose the following sets of coins (since Q1 and Q2 are the same coin value, we can say Q for any instance where either one of these coins would be a possibility): DP, DQ, PQ, Q1Q2. This gives us 4 possible sums (11 cents, 35 cents, 26 cents, or fifty cents.)

If you choose three coins, you could choose the following sets of coins (since Q1 and Q2 are the same coin value, we can say Q for any instance where either one of these coins would be a possibility): DPQ, DQ1Q2, PQ1Q2. This gives us 3 possible sums (36 cents, 60 cents, or 51 cents).

If you choose four coins, you can only choose DPQ1Q2. This gives us 1 possible sum (61 cents.)

Therefore, the total number of different sums of coins you can form is 3 + 4 + 3 + 1 = 11 different sums of money can be produced.

c) The reason that the answers in part (a) and part (b) are not the same is because Q1 and Q2 are treated as two different coins in part (a) whereas they are effectively treated as the same coin in part (b). The reason they are treated as one coin is because they both contribute the same amount to the sum of money.

Read more about Counting Problems at; brainly.com/question/13875198

#SPJ1

3 0
1 year ago
HELP I HAVE TEN MINUTES!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Flura [38]
No


Explanation

Do it on your own
8 0
2 years ago
What is the variable of 7m + 3​
zlopas [31]
The variable would be m Lol
8 0
2 years ago
Read 2 more answers
A box contains 5 red balls, 6 white balls and 9 black balls. Two balls are drawn at
valina [46]

Answer:

P(Same)=\frac{61}{190}

Step-by-step explanation:

Given

Red = 5

White = 6

Black = 9

Required

The probability of selecting 2 same colors when the first is not replaced

The total number of ball is:

Total = 5 + 6 + 9

Total = 20

This is calculated as:

P(Same)=P(Red\ and\ Red) + P(White\ and\ White) + P(Black\ and\ Black)

So, we have:

P(Same)=\frac{n(Red)}{Total} * \frac{n(Red) - 1}{Total - 1} + \frac{n(White)}{Total} * \frac{n(White) - 1}{Total - 1}  + \frac{n(Black)}{Total} * \frac{n(Black) - 1}{Total - 1}

<em>Note that: 1 is subtracted because it is a probability without replacement</em>

P(Same)=\frac{5}{20} * \frac{5 - 1}{20- 1} + \frac{6}{20} * \frac{6 - 1}{20- 1}  + \frac{9}{20} * \frac{9- 1}{20- 1}

P(Same)=\frac{5}{20} * \frac{4}{19} + \frac{6}{20} * \frac{5}{19}  + \frac{9}{20} * \frac{8}{19}

P(Same)=\frac{20}{380} + \frac{30}{380}  + \frac{72}{380}

P(Same)=\frac{20+30+72}{380}

P(Same)=\frac{122}{380}

P(Same)=\frac{61}{190}

4 0
3 years ago
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