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Mumz [18]
2 years ago
5

How do I solve for x plzzzz help

Mathematics
1 answer:
jeka942 years ago
7 0

Answer:

x = -13

Step-by-step explanation:

Find LCM for 5 , 10.  LCM of 5 , 10= 10

\frac{x-3}{5}-\frac{x+1}{10}= -2\\\\\frac{(x-3)*2}{5*2}-\frac{x+1}{10}=-2\\\\\\\frac{2x-6}{10}-\frac{x+1}{10}\\\\\\\frac{2x - 6 - (x+1)}{10}= -2\\\\\\\frac{2x - 6 - x -1}{10} = -2\\\\\\\frac{2x -x - 6 - 1}{10} = -2 \\\\\\\frac{x - 7}{10} = -2\\\\\

Cross multiply,

x - 7 = -2 * 10

x - 7 = -20

     x = -20 + 7

    x = -13

               

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Step-by-step explanation:

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2 years ago
Fabulous fitness charges a one time fee of $90 and an additional $25 per month wonder weights charges a one time fee of only $ 2
Aleksandr-060686 [28]

Answer:

9 months

Step-by-step explanation:

Given

Fabulous fitness charges a one-time fee of $90 and an additional $25 per month

let the number of months be x

and the total charges be y

y= 25x+90---------1

Wonder weights charges a one-time fee of only $ 27 dollars but charge an additional $32 each month

let the number of months be x

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Required

The value of x, that is the number of months

Step two:

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6 0
2 years ago
Suppose heights of seasonal pine saplings are normally distributed and have a known population standard deviation of 17 millimet
Mumz [18]

Answer:

296.693\leq x\leq 319.307

Step-by-step explanation:

The confidence interval for the population mean x can be calculated as:

x'-z_{\alpha /2}\frac{s}{\sqrt{n} } \leq x\leq x'+z_{\alpha /2}\frac{s}{\sqrt{n} }

Where x' is the sample mean, s is the population standard deviation, n is the sample size and z_{\alpha /2} is the z-score that let a proportion of \alpha /2 on the right tail.

\alpha is calculated as: 100%-99%=1%

So, z_{\alpha/2}=z_{0.005}=2.576

Finally, replacing the values of x' by 308, s by 17, n by 15 and z_{\alpha /2} by 2.576, we get that the confidence interval is:

308-2.576\frac{17}{\sqrt{15} } \leq x\leq 308+2.576\frac{17}{\sqrt{15} }\\308-11.307 \leq x\leq 308+11.307\\296.693\leq x\leq 319.307

6 0
3 years ago
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