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aleksklad [387]
4 years ago
13

Robert plans to make a box-and-whisker plot of the following set of data. 27, 14, 46, 38, 32, 18, 21 Find the lower quartile, th

e median, and the upper quartile of the set?

Mathematics
1 answer:
mr Goodwill [35]4 years ago
3 0
Look at the attachment.

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What is the value of w?
solniwko [45]
If you have a no solution choice that will be your answer because w+34=w-22 
you would subtract w from both sides to get 34=-22 and that's not true therefore it would be a no solution equation.
7 0
3 years ago
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PLEASE HELP ME WITH THIS QUESTION ASAP
lana [24]

Answer:

2x + 5y = -4

Step-by-step explanation:

The standard form for linear equations in two variables is Ax+By=C. For example, 2x+3y=5 is a linear equation in standard form. When an equation is given in this form, it's pretty easy to find both intercepts (x and y). This form is also very useful when solving systems of two linear equations.

3 0
3 years ago
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Which equation has solutions of 6 and -6? x2 – 12x + 36 = 0 x2 + 12x – 36 = 0 x2 + 36 = 0 x2 – 36 = 0
Svetlanka [38]

<u>Answer:</u>

The equation that has solutions 6 and -6 is x^2 - 36 = 0

<u>Solution:</u>

We have to find which equation has the solutions 6 and -6.

We have been given three equations.

x^{2}-12 x+36=0  --- eqn 1

x^{2}+12 x-36=0 -- eqn 2

x^{2}-36=0  ---- eqn 3

The 6 and -6 to satisfy any of these equations they have to be the roots of the equation.

This means that when we substitute 6 and -6 in any of the equations and then solve them the answer on simplification should be 0.

This condition should individually be satisfied by both 6 and -6 for any one of the equations.

Now let us try and substitute 6 and -6 in eq1.

Now, substituting 6 in eq1.

62-12×6+36=0

Now we simply the equation to check is the LHS is equal to the RHS of the equation.  

LHS:

72-72=0

RHS:  0  

Since LHS=RHS it is the root of the equation.

Now we check if -6 satisfies eq1.

-62-12×-6+36=0

LHS:

72+72=144

RHS:  0

Hence LHS is not equal to RHS, -6 is not the root of eq1.

Similarly we check for eq2  

Checking for 6 and -6 we get

LHS is not equal to RHS hence this does not satisfy eq2.

Now in the same way we check for eq3

LHS=RHS for both 6 and -6 hence they are the solutions for eq3.

Hence the equation that has solutions 6 and -6 is x^2 - 36 = 0

3 0
3 years ago
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Write the equation of the line in slope intercept form.
skelet666 [1.2K]
Y = -2/3x + 3
That’s the equation
4 0
3 years ago
Need help with a and b parts
Masja [62]

Answer:

a)  10/27 or 0.37037

b) -8/7

Step-by-step explanation:

a)  Enter x=-5 into both equations and divide h(-5)/g(-5).  That yields 10/27 or 0.37037037037 . . .

b) g(x) cannot be 0 (can't divide by 0.  That would occur if x = -8/7, so it (-8/7) is not in the domain.

6 0
3 years ago
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