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max2010maxim [7]
4 years ago
7

If (x-2) (3x-2) = 0, what are all the possible values of x?

Mathematics
2 answers:
dalvyx [7]4 years ago
6 0
By setting both equations to zero and getting x alone you find your anwer
x-2=0        3x-2=0
 +2  +2         +2  +2
x=2           3x=2
                 /3  /3
                   x=2/3
Allisa [31]4 years ago
5 0
<span>(x-2)(3x-2) = 0
F1      F2

F1 means factor 1 and F2 means Factor 2 (of multiplication)

Then you do x-2 = 0 and 3x-2 = 0

x-2 = 0
Move the common factor to the right with changed sign

x = 2

3x-2 = 0
3x = 2

</span>\frac{3}{3} x =  \frac{2}{3}
<span>
x = 2/3

So your answer is B</span>
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A bag of flour weighed 2.25 kilograms 0.725 kilograms are used in a recipe how many kilograms of flour are left
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The distribution of SAT II Math scores is approximately normal with mean 660 and standard deviation 90. The probability that 100
gayaneshka [121]

Using the <em>normal distribution and the central limit theorem</em>, it is found that there is a 0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

  • It measures how many standard deviations the measure is from the mean.
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
  • By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation s = \frac{\sigma}{\sqrt{n}}.

In this problem:

  • The mean is of 660, hence \mu = 660.
  • The standard deviation is of 90, hence \sigma = 90.
  • A sample of 100 is taken, hence n = 100, s = \frac{90}{\sqrt{100}} = 9.

The probability that 100 randomly selected students will have a mean SAT II Math score greater than 670 is <u>1 subtracted by the p-value of Z when X = 670</u>, hence:

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{670 - 660}{9}

Z = 1.11

Z = 1.11 has a p-value of 0.8665.

1 - 0.8665 = 0.1335.

0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

To learn more about the <em>normal distribution and the central limit theorem</em>, you can take a look at brainly.com/question/24663213

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