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Finger [1]
3 years ago
10

Emily quiz scores for this semester were 82,73,73,91 and 85 what is the

Mathematics
1 answer:
Elanso [62]3 years ago
3 0
The median is 82
The mode is 73
The mean is 80.8
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How do i factor the expressions 4w-16?
igomit [66]
The answer to this would be 4(w-4)
7 0
3 years ago
Solve the equation A = bh for b.
kotegsom [21]

Answer:

b=A/h

Step-by-step explanation:

To isolate b, you need to divide both sides by h. Therefore:

A=bh

b=A/h

Hope this helps!

8 0
3 years ago
Read 2 more answers
20 points <br><br><br> pppppplease help me!!!!!
pogonyaev

Answer:

dont know if this is correct

5 0
2 years ago
write the linear equation from the given information in slope intercept form passing through (-2,8) and perpendicular to y=2x+5​
Leto [7]

Answer:

y  =  -\frac{1}{2} x + 9

Step-by-step explanation:

Given parameters:

  Coordinates = (-2,8)

Equation of the line  y  = 2x + 5;

Unknown:

Linear equation of a line perpendicular = ?

Solution:

From the given expression, a line perpendicular will have a negative inverse of the slope given.

           y  = 2x + 5

Equation of a line  is y = mx + c;

y and x are the coordinates

  m is the slope

   c is the intercepts

    y  = 2x + 5 ;

 The slope is 2;

  A line perpendicular will have a slope of -\frac{1}{2}  

 Since y = 8 and x  = -2, let us find c;

     8  = 2 x -\frac{1}{2}  + c

    8 = -1 + c

     c  = 8+1 = 9

The equation of the line is;

       y  =  -\frac{1}{2} x + 9

7 0
2 years ago
Suppose you have a water-balloon launcher. The balloon is 3 ft high when it leaves the launcher. Use the equation 0 = ?16t2 + 38
mario62 [17]

Answer:

t = 2.5 s

Step-by-step explanation:

Suppose you have a water-balloon launcher. The balloon is 3 ft high when it leaves the launcher. Its equation is :

16t^2 + 38.8t + 3 =0

The above equation is a quadratic equation. The general equation is :

ax^2+bx+c=0

Here, a = 16, b = 38.8 and c = 3

The solution of quadratic equation is given by :

x=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}\\\\t=\dfrac{-38.8\pm \sqrt{(38.8)^2-4\times (-16)\times 3} }{2\times (-16)}\\\\t=\dfrac{-38.8+\sqrt{(38.8)^{2}-4\times-16\times3}}{-2\times16}, \dfrac{-38.8-\sqrt{(38.8)^{2}-4\times-16\times3}}{-2\times16}\\\\t=-0.075\ s,2.5\ s

So, at t = 2.5 s the balloon is in air.

8 0
3 years ago
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