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Tems11 [23]
3 years ago
7

Gerardo made the following scores on his first semester science tests: 80, 85, 85, 87, 97, 100. Which measure would he use in su

mmarizing his scores to give the most favorable impression of his performance? mean median mode range
Mathematics
1 answer:
Elza [17]3 years ago
4 0

The mean is 89 the median is 86 the mode is 85 Hope this helps!

(technically the mean is like 88.7 so I just rounded up to get 89)

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Write the expression in simplest form using only positive exponents (x^2/3y^-1/2)^-6
hoa [83]
\left( \cfrac{x^2}{3y^{- \frac{1}{2} }} \right)^{-6}=\left( \cfrac{3y^{- \frac{1}{2} }}{x^2} \right)^{6}=\left( \cfrac{3}{x^2y^{ \frac{1}{2} }} \right)^{6}= \cfrac{3^6}{x^{2*6}y^{ \frac{1}{2}*6 }} = \cfrac{729}{x^{12}y^3}
5 0
3 years ago
Degrees =x<br><br><br> degrees =y<br><br><br> degrees =z<br>Please help
suter [353]

Answer:

x = 64

z = 98

y= 18

Step-by-step explanation:

x = 64

64+z+18 = 180

z = 180-82 = 98

y= 18

6 0
3 years ago
3xy=9 is it direct variation or inverse variation or neither
steposvetlana [31]

3xy = 9


y =  \dfrac 9 {3x}


y  = \dfrac 3 x


y varies inversely with x


Answer: inverse variation



5 0
3 years ago
A function, F(x), is shown below.
Maksim231197 [3]

Answer:

  range of f(x) = [-4, -2) ∪ [2, 8)

  a+b+c+d = -4

Step-by-step explanation:

The graph is attached. The range is the vertical extent of the function. It is defined at f(0) = -4 and f(2) = 2.

The limits f(2-) and f(4-) are -2 and 8, respectively, so the graph has open circles there. These are the ends of the two half-open intervals that make up the range of the function.

The portion of the graph in the domain [4, 7) is included in the range [2, 8), so no special treatment is needed for that piece of the function.

7 0
3 years ago
Need math help asap please! :/ thanks
Dominik [7]

The rule to remember about generating the perpendicular family to a line is we swap the coefficients on and x and y, remembering to negate one of them. Then the constant is set directly from the intersecting point.


So we have


y = 3x + 2


-3x + 1y = 2


Swapping and negating gets the perpendiculars; the constant is as yet undetermined.


1x + 3y = constant


Since we want to go through (0,2), we could have just written


x + 3y = 0 + 3(2) = 6


3y = -x + 6


y = (-1/3) x + 2


Third choice


5 0
2 years ago
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