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Nana76 [90]
3 years ago
13

Solve the absolute value equation (picture listed below)

Mathematics
1 answer:
STALIN [3.7K]3 years ago
3 0

Answer:

17) x = 0, -6

18) x = 2, -4

Step-by-step explanation:

Isolate the absolute value, then split into 2 equations and solve both

example: |x + 1| = 2

x + 1 = 2, x + 1 = -2

x = 1, -3

------------------------------------

17) |x + 3| - 1 = 2

isolate the absolute value:

|x + 3| = 3

x + 3 = 3, x + 3 = -3

x = 0, -6

-----------------------------------------

18) 3|x + 1| = 9

isolate absolute value:

|x + 1| = 3

x + 1 = 3, x + 1 = -3

x = 2, -4

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Step-by-step explanation:

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The scores on the LSAT are approximately normal with mean of 150.7 and standard deviation of 10.2. (Source: www.lsat.org.) Queen
faltersainse [42]

Answer:

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Step-by-step explanation:

1) 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".  

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Let X the random variable that represent the  scores on the LSAT of a population, and for this case we know the distribution for X is given by:

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We want to find a value a, such that we satisfy this condition:

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P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.35 of the area on the left and 0.65 of the area on the right it's z=-0.385. On this case P(Z<-0.385)=0.35 and P(Z>-0.385)=0.65

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

Z=-0.385

And if we solve for a we got

a=150.7 -0.385*10.2=146.773

So the value of height that separates the bottom 35% of data from the top 65% (Or the 35 percentile) is 146.7.  

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