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3 years ago

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Alina fully simplifies this polynomial and then writes it in standard form. xy2 – 2x2y + 3y3 – 6x2y + 4xy2 If Alina wrote the la

st term as 3y3, which must be the first term of her polynomial in standard form? xy2 5xy2 –8x2y –2x2y

2 answers:

ad-work [718]3 years ago

6 0

Answer:

$-4x^2y$

Step-by-step explanation:

Alina wrote the polynomial

$2x^2y+3y^3-6x^2y+4xy^2$

She wrote $3y^3$ as the last term

We will solve the like terms and get its simplified form which is

$-4x^2y+4xy^2+3y^3$

Hence, the first term according to Alina standard form would be $-4x^2y$

Liono4ka [1.6K]3 years ago

3 0

I think the answer is -8x2y

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To find the answer you'd first add up how many patients you have:

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3. Ramona drank 1 5/8 liters of water during soccer practice. She drank 1 1/4

VashaNatasha [74]

Answer:

The expression x = $\frac{13}{8}+\frac{5}{4}$ could be used to determine the amount of water drank altogether.

Step-by-step explanation:

Given that:

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Let,

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The expression x = $\frac{13}{8}+\frac{5}{4}$ could be used to determine the amount of water drank altogether.

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Step-by-step explanation:

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Alex787 [66]

Answer:

There is a 0.058% probability that this student will pass the examination.

Step-by-step explanation:

For each question, there are only two possible outcomes. Either it is correct, or it is wrong. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

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There are 12 questions, so $n = 12$ .

The student guesses each question. There are five possible answers, only one which is correct, so $p = 0.2$ .

What is the probability that a student who guesses at random on each question will pass the examination?

This is $P(X \geq 8)$

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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$P(X = 9) = C_{12,9}.(0.2)^{9}.(0.8)^{3} = 0.000058$

$P(X = 10) = C_{12,10}.(0.2)^{10}.(0.8)^{2} = 0.000004$

$P(X = 11) = C_{12,11}.(0.2)^{11}.(0.8)^{1} = 0.0000002$

$P(X = 12) = C_{12,12}.(0.2)^{12}.(0.8)^{0} = 0.000000004$

So

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.000519 + 0.000058 + 0.000004 + 0.0000002 + 0.000000004 = 0.00058$

There is a 0.058% probability that this student will pass the examination.

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