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statuscvo [17]
2 years ago
10

HELP ME SOLVE HURRY I WILL MARK BRAINLIEST THANKS

Mathematics
1 answer:
adelina 88 [10]2 years ago
5 0

Answer:

your y intercept is 90 and the equation would be y= 90 and x is unknown

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3 years ago
To estimate a population mean, the sample size needed to provide a margin of error of 3 or less with a 95% confidence when the p
Over [174]

Answer:

The minimum sample needed to provide a margin of error of 3 or less is 52.

Step-by-step explanation:

The confidence interval for population mean (<em>μ</em>) is:

\bar x\pm z_{\alpha /2}\frac{\sigma}{\sqrt{n}}

The margin of error is:

MOE= z_{\alpha /2}\frac{\sigma}{\sqrt{n}}

<u>Given:</u>

MOE = 3

<em>σ </em>= 11

The critical value for 95% confidence interval is: z_{\alpha /2}=1.96

**Use the <em>z</em>-table for critical values.

Compute the sample size (<em>n</em>) as follows:

MOE= z_{\alpha /2}\frac{\sigma}{\sqrt{n}}\\3=1.96\times \frac{11}{\sqrt{n}}\\n=(\frac{1.96\times11}{3} )^{2}\\=51.65\\\approx52

Thus, the minimum sample needed to provide a margin of error of 3 or less is 52.

7 0
3 years ago
If x+1/x= 3, then prove that m^5+1/m^5= 123
alina1380 [7]

9514 1404 393

Explanation:

We can start with the relations ...

  \displaystyle\left(x+\frac{1}{x}\right)^3=\left(x^3+\frac{1}{x^3}\right)+3\left(x+\frac{1}{x}\right)\\\\\left(x+\frac{1}{x}\right)^5=\left(x^5+\frac{1}{x^5}\right)+5\left(x^3+\frac{1}{x^3}\right)+10\left(x+\frac{1}{x}\right)\\\\\textsf{From these, we can derive ...}\\\\x^5+\frac{1}{x^5}=\left(x+\frac{1}{x}\right)^5-5\left(\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)\right)-10\left(x+\frac{1}{x}\right)

  \displaystyle x^5+\frac{1}{x^5}=\left(x+\frac{1}{x}\right)^5-5\left(x+\frac{1}{x}\right)^3+5\left(x+\frac{1}{x}\right)\right)\\\\x^5+\frac{1}{x^5}=3^5 -5(3^3)+5(3)\\\\=((3^2-5)3^2+5)\cdot3=(4\cdot9+5)\cdot3=(41)(3)\\\\=\boxed{123}

7 0
3 years ago
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