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2 years ago

(4x2y3 + 2xy2 – 2y) – (–7x2y3 + 6xy2 – 2y) i need help solving

Mathematics

1 answer:

lubasha [3.4K]2 years ago

5 0

Answer:

$\huge \boxed{11x^2 y^3-4xy^2 }$

Step-by-step explanation:

$(4x^2y^3 + 2xy^2 - 2y) - (-7x^2 y^3 + 6xy^2 -2y)$

Distribute the negative sign to the terms in the bracket.

$(4x^2y^3 + 2xy^2 - 2y) +7x^2 y^3 - 6xy^2 +2y$

Grouping like terms.

$(4x^2y^3 +7x^2 y^3) + (2xy^2 - 6xy^2 ) +( - 2y +2y)$

Combining like terms.

$(11x^2 y^3) + (-4xy^2 ) +(0)$

$11x^2 y^3-4xy^2$

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Anettt [7]

Answer:

Choice C: approximately 121 green beans will be 13 centimeters or shorter.

Step-by-step explanation:

What's the probability that a green bean from this sale is shorter than 13 centimeters?

Let the length of a green bean be $X$ centimeters.

$X$ follows a normal distribution with

mean $\mu = 11.2$ and
standard deviation $\sigma = 2.1$ .

In other words,

$X\sim \text{N}(11.2, 2.1^{2})$ ,

and the probability in question is $X \le 13$ .

Z-score table approach:

Find the z-score of this measurement:

$\displaystyle z= \frac{x-\mu}{\sigma} = \frac{13-11.2}{2.1} = 0.857143$ . Closest to 0.86.

Look up the z-score in a table. Keep in mind that entries on a typical z-score table gives the probability of the left tail, which is the chance that $Z$ will be less than or equal to the z-score in question. (In case the question is asking for the probability that $Z$ is greater than the z-score, subtract the value from table from 1.)

$P(X\le 13) = P(Z \le 0.857143) \approx 0.8051$ .

"Technology" Approach

Depending on the manufacturer, the steps generally include:

Locate the cumulative probability function (cdf) for normal distributions.
Enter the lower and upper bound. The lower bound shall be a very negative number such as $-10^{9}$ . For the upper bound, enter $13$
Enter the mean and standard deviation (or variance if required).
Evaluate.

For example, on a Texas Instruments TI-84, evaluating $\text{normalcdf})(-1\text{E}99,\;13,\;11.2,\;2.1 )$ gives $0.804317$ .

As a result,

$P(X\le 13) = 0.804317$ .

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Assume that the length of green beans for sale are independent of each other. The probability that each green bean is shorter than 13 centimeters is constant. As a result, the number of green beans out of 150 that are shorter than 13 centimeters follow a binomial distribution.

Number of trials $n$ : 150.
Probability of success $p$ : 0.804317.

Let $Y$ be the number of green beans out of this 150 that are shorter than 13 centimeters. $Y\sim\text{B}(150,0.804317)$ .

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To determine the common ratio of a geometric sequence. You just need to divide any two consecutive terms on it. You can see below that all of them have the same quotient.

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