15x-6x+5xsquared -6x cubed - 2x squared +2(6x-2) please help me!!! (Show work)

1 answer:

6 0

$15x-6x+5x^2-6x^3-2x^2+2(6x-2)\\\\=15x-6x+5x^2-6x^3-2x^2+12x-4\\\\=(-6x^3)+(5x^2-2x^2)+(15x-6x+12x)+(-4)\\\\=-6x^3+3x^2+21x-4$

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A soccer team won only 10 of their first 30 games. How many of their remaining 20 games will they have to win to have a .500 rec

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Answer: 5 games

Step-by-step explanation: A .500 record means that the soccer team won half of their games ( that's how I interpret it). Since they are going to play 30 games, they need to win half of 30 or 15 games to have that record. 15-10 = 5. So, they need to win 5 more games to have a .500 record.

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What is the z-score of a newborn who weighs 4,000 g?.

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The z-score of the considered newborn having 4,000 g weight is $\dfrac{4000 - \mu}{\sigma}$

<h3>How to get the z scores?</h3>

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.

If we have

$X \sim N(\mu, \sigma)$

(X is following normal distribution with mean $\mu$ and standard deviation $\sigma$ )

then it can be converted to standard normal distribution as

$Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)$

(Know the fact that in continuous distribution, probability of a single point is 0, so we can write

$P(Z \leq z) = P(Z < z) )$

Also, know that if we look for Z = z in z tables, the p value we get is

$P(Z \leq z) = \rm p \: value$

For the considered case, the statement in the problem is missing the mean weight of the newborn and standard deviation of the data set of weights of newborn.

Thus, for them, we can consider variables in place of their actual values.

Let we have: