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Olenka [21]
1 year ago
6

Consider the system of inequalities and its graph. y ≤ –0.75x y ≤ 3x – 2 In which section of the graph does the actual solution

to the system lie? 1 2 3 4

Mathematics
2 answers:
-Dominant- [34]1 year ago
7 0

Answer:

The section shows below represents the solution of the system.

Step-by-step explanation:

Given system of inequalities,

y ≤ -0.75x,

y ≤ 3x - 2

Graphing y ≤ -0.75x :

Since, the related equation of y ≤ -0.75x is y = 0.75x

If x = 0, 1,

y = 0, 0.75,

Thus, join the points (0, 0) and (1, 0.75) in coordinate plane,

0 ≤ -0.75(0) ( true )

So, the shaded region will be contain the origin.

Graphing y ≤ 3x - 2 :

Since, the related equation of y ≤ 3x - 2 is y = 3x - 2

x = 0, y = - 2

y = 0, x = 2/3,

Thus, join the points (0, -2) and (2/3, 0) in coordinate plane,

0 ≤ 3(0) - 2 ( false )

So, the shaded region won't contain the origin.

Also, ' ≤ ' represents the solid line,

By the above explanation we can make the feasible region that shows the solution of the system ( shown below )

lora16 [44]1 year ago
3 0
We need to assign a value for x to check the possible values of y. 1st inequality: y < -0.75x X = - 1 ; y < -0.75(-1) ; y < 0.75 possible coordinate (-1,0.75) LOCATED AT THE 2ND QUADRANT X = 0 ; y < -0.75(0) ; y < 0 possible coordinate (0,0) ORIGIN X = 1 ; y < -0.75(1) ; y < -0.75 possible coordinate (1,-0.75) LOCATED AT THE 4TH QUADRANT 2nd inequality: y < 3x -2 X = -1 ; y < 3(-1) – 2 ; y < -5 possible coordinate (-1,-5) LOCATED AT THE 4TH QUADRANT X = 0 ; y < 3(0) – 2 ; y < -2 possible coordinate (0,-2) LOCATED AT THE 4TH QUADRANT X = 1 ; y < 3(1) – 2 ; y <<span> 1 possible coordinate (1,1) LOCATED AT THE 1ST QUADRANT The actual solution to the system lies on the 4TH QUADRANT. </span>
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Using the value of a₁ and d, we can simplify the expression as:

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8 0
2 years ago
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The United States Bureau of Labor Statistics (BLS) conducts the Quarterly Census of Employment and Wages (QCEW) and reports a va
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Answer:

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  The  probability is  P(\= X < 33) = 0.8531

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  The  probability is  P(\= X > 30) = 0.3520

Step-by-step explanation:

From the  question we are told that

     The population mean is  \mu =  28.29

      The standard deviation is \sigma  =  33.493

       The sample size is  n  = 56

Generally the standard error for the  sample  mean (\= x ) is mathematically evaluated as

        \sigma _{\=x} =  \frac{\sigma}{\sqrt{n} }

substituting values  

       \sigma _{\=x} =  \frac{33.493}{\sqrt{56} }

      \sigma _{\=x} = 4.48

Apply central limit theorem[CLT] we have  that

        P(\= X < 33) =  [z <  \frac{33 -  \mu }{\sigma_{\= x}} ]

substituting values

       P(\= X < 33) =  [z <  \frac{33 -  28.29 }{4.48} ]

       P(\= X < 33) =  [z <  1.05 ]

From the z-table  we have that  

       P(\= X < 33) = 0.8531

For the second question

    Apply central limit theorem[CLT] we have  that

    P(\= X > 30 ) =  [z >  \frac{30 -  \mu }{\sigma_{\= x}} ]

substituting values

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From the z-table  we have that  

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Thus  

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Answer:

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Step-by-step explanation:

These are three prediction that Keisha can make based on the report that said 6 of 300 clocks tested weren't working.

Base on that information, Keisha can calculate an experimental probability, dividing <em>clocks that don't work properly </em>by <em>the total amount of clocks</em><em>:</em>

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