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artcher [175]
3 years ago
13

30. Solve the equation below. 2 X- - 5 = 7 PLEASE HELP !!

Mathematics
2 answers:
nalin [4]3 years ago
4 0

9514 1404 393

Answer:

  x = 18

Step-by-step explanation:

Add the opposite of the constant (-5), and multiply by the inverse of the coefficient of x (2/3).

  \dfrac{2}{3}x -5=7\\\\\dfrac{2}{3}x=12 \qquad\text{add 5}\\\\x=\dfrac{3}{2}\cdot 12 \qquad\text{multiply by $\dfrac{3}{2}$}\\\\ \boxed{x=18}

Serga [27]3 years ago
4 0

Answer:

2/3x -5=7

x=18

Step-by-step explanation:

start by adding each side by 5

after doing so you should have

2/3x =12

multiply each side now my 3/2

the fractions should cancel out and leave x by itself

multiply (12)3/2

gives you 18.

x=18

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Define z_alpha to be a z-score with an area of alpha to the right. For Example: z_0.10 means P(Z &gt; z_0.10) = 0.10. We would a
Reptile [31]

Answer:

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For this case we want a quantile that accumulates 0.025 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

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c) For this case we want to find a value of z that satisfy:

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And we can use the following excel code:

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Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Part a

P(-z_0.025 < Z < z_0.025)

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"=NORM.INV(0.025,0,1)"

"=NORM.INV(0.025,0,1)"

And for this case the two values are :z_{crit}= \pm 1.96

Part b

P(-z_{\alpha/2} < Z < z_{\alpha/2})

For this case we want a quantile that accumulates \alpha/2 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(alpha/2,0,1)"

"=NORM.INV(alpha/2,0,1)"

Part c

For this case we want to find a value of z that satisfy:

P(Z > z_alpha) = 0.05.

And we can use the following excel code:

"=NORM.INV(0.95,0,1)"

And we got z_{\alpha/2}=1.64

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