$\dfrac{3(2x^4y^5)^2}{(6x^3y^3)^6}\qquad\text{use}\ (ab)^n=a^nb^n\ \text{and}\ (a^n)^m=a^{nm}\\\\=\dfrac{3(2^2)(x^4)^2(y^5)^2}{6^6(x^3)^6(y^3)^6}=\dfrac{3(4)x^8y^{10}}{46656x^{18}y^{18}}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=\dfrac{12x^8y^{10}}{46656x^8x^{10}y^{10}y^8}=\dfrac{1}{3888x^{10}y^8}$

7 0

3 years ago

The perimeter of a regulation singles tennis court is 210 feet and the length is 57 feet less than five times the width.

Marianna [84]

Answer:

Length = 24 and Width = 81

Step-by-step explanation:

Let Width be W

then Length = W-57

Perimeter of rectangle = 2(Length + Width)

=2 (W-57+W)= 210

4W- 114= 210

4W= 324

W= 81

Then Length = 81-57 = 24

4 0

3 years ago

Please help i’m in a hurry

Wittaler [7]

Answer:

the answer should be 18

Step-by-step explanation:

this isnt exactly hard when you get the first part with ST beeing 36 then you can just halve that and the answer is 18

6 0

3 years ago

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

vagabundo [1.1K]

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:

X = weights of newborns for some considered system (in grams)
$\mu$ = mean weight of newborns (in grams)
$\sigma$ = standard deviation of the data set of weights of newborns

Then, the z-score for X = 4000 is

$z = \dfrac{x - \mu}{\sigma} = \dfrac{4000 - \mu}{\sigma}$

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

Learn more about z-scores here:

brainly.com/question/13299273

3 0

2 years ago