<img src="https://tex.z-dn.net/?f=%281%20x%5E%7B2000%7D%29%281%20x%29%5E%7B1998%7D%3D%282x%29%5E%7B1999%7D" id="TexFormula1" tit

All categories

3 years ago

le="(1 x^{2000})(1 x)^{1998}=(2x)^{1999}" alt="(1 x^{2000})(1 x)^{1998}=(2x)^{1999}" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

Maksim231197 [3]3 years ago

7 0

Hello,
$x^{2000}*x^{1998}=2^{1999}*x^{1999}\\
==\textgreater\ x^{2000+1998}=2^{1999}*x^{1999}\\
==\textgreater\ x^{2000+1998-1999}=2^{1999}\\
==\textgreater\ x^{1999}=2^{1999}\\
==\textgreater\ x=2


$

You might be interested in

The local high school’s cheerleading squad is selling snow cones during lunch to raise money for a national cheerleading competi

Artemon [7]

Answer:

I can just imagine some lady with big jugs leaning over the snow cones and saying in a extremely smexual voice "would you like these huge snow cones baby?", also its roughly 4 cm wide

Step-by-step explanation:

5 0

3 years ago

A consumer group selected 100 different airplanes at random from each of two large airlines. The mean seat width for the 100 air

Novay_Z [31]

Answer:

Decreasing the sample size decreases the width of the confidence interval.

Step-by-step explanation:

-Let s be the sample standard deviation and $\bar X$ be the sample mean.

#We calculate the 95% confidence interval for the sample size 100 as below:

$\bar X\pm z_{0.05/2}\frac{s}{\sqrt{n}}\\\\\\=\bar X\pm 1.960\times\frac{s}{\sqrt{50}}\\\\\\=\bar X\pm 0.1960s$

#we then calculate the confidence interval of sample size =50 and of similar mean and standard deviation:

$\bar X\pm z\frac{s}{\sqrt{n}}\\\\\\=\bar X\pm z_{0.05/2}\times\frac{s}{\sqrt{n}}\\\\\\=\bar X\pm1.960\times \frac{s}{\sqrt{50}}\\\\\\\\=\bar X\pm0.2772s$