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pentagon [3]
4 years ago
6

The summary statistics for all students that took the SAT at Jones High School are shown. Four sample groups of 10 students are

Mathematics
2 answers:
8090 [49]4 years ago
6 0

Answer:

D)

Step-by-step explanation:

i got it corect on the test

Doss [256]4 years ago
4 0

Answer:

i got the wrong answer when i picked d and the answer is group b

Step-by-step explanation:

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Hi! help is greatly appreciated!
devlian [24]

Answer:

answer is 3

because first we add 4+2=6

the sum of the line is 6 then we will got

x+3=6

x=6-3

x=3

Step-by-step explanation:

hope my answer is helpful to you

plz mark me as brainlist

6 0
4 years ago
Read 2 more answers
Please help. I don’t understand what to do
11111nata11111 [884]

\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=23.9\\ h=100 \end{cases}\implies V=\cfrac{\pi (23.9)^2(100)}{3} \\\\\\ V=\cfrac{57121\pi }{3}\implies V\approx 59816.97\implies \stackrel{\textit{rounded up}}{V=59817} \\\\[-0.35em] ~\dotfill

now, for the second one, we know the diameter is 10, thus its radius is half that or 5.

\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=5\\ V=225 \end{cases}\implies 225=\cfrac{\pi (5)^2 h}{3}\implies 225=\cfrac{25\pi h}{3} \\\\\\ \cfrac{225}{25\pi }=\cfrac{h}{3}\implies \cfrac{9}{\pi }=\cfrac{h}{3}\implies \cfrac{27}{\pi }=h\implies 8.59\approx h\implies \stackrel{\textit{rounded up}}{8.6=h}

6 0
4 years ago
What is the distance between the ordered pairs (-5.5, -3) and (6.5, -3)
motikmotik

Step-by-step explanation:

√(-5.5-6.5)²+(-3+3)²

=√144

=12

4 0
3 years ago
Read 2 more answers
Which describes the cost to produce one item?
Vlada [557]
The answer is either B or D
3 0
3 years ago
Read 2 more answers
Suppose f is a function of one variable that has a continuous second derivative. Show that for any constants a and b, the functi
salantis [7]

Answer:

We can find the solution using chain rule,

For instance, if u(x,y) = f(ax+by), then

u_{x}(x,y) = a f^{'} (ax+by)  which represents derivative of x with respect to x

and then derivative of u_{x}(x,y) with respect to y is

u_{xy}(x,y) = (ab)^2 (f^{''} (ax+by))^2,  

Now, the derivative of u_{x}(x,y) with respect to x, which is the second derivative, which is

u_{xx}(x,y) = a^2 f^{''} (ax+by)

and the derivative u_{y}(x,y) and   u_{yy}(x,y) are

u_{y}(x,y) = b f^{'} (ax+by),

u_{yy}(x,y) = b^2 f^{''} (ax+by)

Finally, the solution of PDE is

u_{xx}(x,y) u_{yy}(x,y) - u_{xy} ^2 (x,y)

= (a^2 f^{''} (ax+by)) (b^2 f^{''} (ax+by)) - (ab)^2 (f^{''} (ax+by))^2

= (ab)^2 (f^{''} (ax+by))^2) - (ab)^2 (f^{''} (ax+by))^2

= 0,

As the PDE is equal to 0, it means the function u(x,y) = f(ax+by) is the solution of the given PDE.

6 0
4 years ago
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