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goldfiish [28.3K]
3 years ago
10

20-%201%29%20%2B%202%20%5Cdiv%20%28x%20-%203%29%28x%20-%201%29" id="TexFormula1" title="1 \div (x - 2) (x - 3) + 1 \div (x - 2)(x - 1) + 2 \div (x - 3)(x - 1)" alt="1 \div (x - 2) (x - 3) + 1 \div (x - 2)(x - 1) + 2 \div (x - 3)(x - 1)" align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
Anika [276]3 years ago
7 0
I don’t know if you need to show work but here is the answer 2(x-2)/x-3
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If 1,000 students take a test that has a mean of 40 minutes, a standard deviation of 8 minutes, and is normally distributed, how
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500 because in a normal distribution 50% of the total falls below the mean 
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3 years ago
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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

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3 0
1 year ago
What is the next number in the series 167, 118, 82
Katyanochek1 [597]

The next number in the series are 57, 41, 37, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2.

<h3>What is a sequence?</h3>

It is defined as the systematic way of representing the data that follows a certain rule of arithmetic.

It is given that:

The series is:

167, 118, 82

A number is a mathematical entity that can be used to count, measure, or name things. For example, 1, 2, 56, etc. are the numbers.

As we know, the prime is divisible by 1 and itself.

The prime numbers before 82 are:

57, 41, 37, 31, 29, 23, 19.. and so on

The next number in series are 57, 41, 37, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2

Thus, the next number in the series are 57, 41, 37, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2.

Learn more about the sequence here:

brainly.com/question/21961097

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1 year ago
Rational root theorem
jok3333 [9.3K]

Answer: a_{n} x^{n} +a_{n-1}x^{n-1}+...+a_{1}x^{1} +a_{0}=0

8 0
3 years ago
How I can solve this
gavmur [86]

Answer:

\boxed{\textrm{B)}\qquad y>4}.

Step-by-step explanation:

We start by subtracting 1 from both sides of the second inequality:

2x > 5 \iff 2x - 1 > 5 - 1 \iff 2x - 1 > 4.

Therefore, we conclude that:

y > 2x - 1 > 4 \implies y > 4.

Therefore, every point that satisfies the system has a y-coordinate greater than 4.

7 0
3 years ago
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