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3 years ago

PLS HELP What is the median of this data set?

Mathematics

1 answer:

balu736 [363]3 years ago

7 0

Answer:

$3.50

Step-by-step explanation:

add up all the x's and distribute them starting from the outside until you get to the middle. if there is an odd number of x's (which there isn't), the last x to be placed will be on the spot that is the median. If there is an even number of x's, and they both land on the same spot, that spot will be the median, and if they are next to each other, the median will be between them.

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PLEASE HELP!!!!!! I CANT FIGURE IT OUT

Dmitriy789 [7]

Answer:8

Step-by-step explanation:

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3 years ago

Please Help Quickly!!! Find the limit if f(x) = x^3

jeka94

Answer:

Option b. 12

Step-by-step explanation:

This exercise asks us to find the derivative of a function using the definition of a derivative.

Our function is $f(x) = x^{3}$ . Therefore:

$f(2+h) = (2+h)^{3}$

$f(2) = (2)^{3} = 8$

Then:

$\lim_{h \to \0} \frac{f(2+h)-f(2)}{h}=\lim_{h \to \0} \frac{(2+h)^{3}-8}{h}$

Expanding:

$\lim_{h \to \0} \frac{(2+h)^{3}-8}{h} =\lim_{h \to \0} \frac{8+ h^{3} +6h(2+h) -8}{h} =\lim_{h \to \0} \frac{h^{3} +6h(2+h)}{h}$

$\lim_{h \to \0} \frac{h^{3}+ 6h(2+h)}{h} =\lim_{h \to \0} h^{2} + 6(2+h)$

Now, if x=0:

$\lim_{h \to \0} \frac{f(2+h)-f(2)}{h} = (0)^{2} +6(2+0) = 12$

4 0

3 years ago

Circle the equation that does not belong and explain why?

Andru [333]

It's either b or c I am not sure

7 0

4 years ago

Find the standard form of the equation of the parabola with a focus at (-2, 0) and a directrix at x = 2.

krok68 [10]

Answer:

$y^2=\frac{1}{8}x$

Step-by-step explanation:

The focus lies on the x axis and the directrix is a vertical line through x = 2. The parabola, by nature, wraps around the focus, or "forms" its shape about the focus. That means that this is a "sideways" parabola, a "y^2" type instead of an "x^2" type. The standard form for this type is

$(x-h)=4p(y-k)^2$

where h and k are the coordinates of the vertex and p is the distance from the vertex to either the focus or the directrix (that distance is the same; we only need to find one). That means that the vertex has to be equidistant from the focus and the directrix. If the focus is at x = -2 y = 0 and the directrix is at x = 2, midway between them is the origin (0, 0). So h = 0 and k = 0. p is the number of units from the vertex to the focus (or directrix). That means that p=2. We fill in our equation now with the info we have:

$(x-0)=4(2)(y-0)^2$