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3 years ago

8

PLEASE HELP!!! These dot plots show the ages in years) for a sample of two types of fish.

1 answer:

worty [1.4K]3 years ago

7 0

<h3>Answers: A and B</h3>

Explanation:

The median is the center of the distribution. We see that the center of the shark's distribution is to the left compared to the koi's center. Therefore, the shark's median age is smaller. Choice A is one of the answers.

The spread is exactly what it sounds like: how spread out the data is. Mathematically we use the standard deviation, or sometimes the range, to find out how spread out things are. The koi distribution is more spread out. The shark's data is more clumped together. This is why choice B is the other answer.

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La temperatura a una distancia r del centro de una lámina está dada por T=40 (r2?2r) . La variación instantánea de la temperatur

PilotLPTM [1.2K]

For this, the first thing to do is to assume that the function of temperature with respect to r is written in one of the following ways:

Way 1:

$T = 40 (r ^ 2 + 2r)$

Way 2:

$T = 40 (r ^ 2-2r)$

To find the instant variation we must find the derivative of the temperature with respect to the distance r.

We have then:

For function 1:

$\frac {dT} {dr} = 40 \frac {d ((r ^ 2 + 2r))} {dr}\\$

$\frac {dT} {dr} = 40 (2r + 2)$

Rewriting

$\frac {dT} {dr} = 80r + 80$

For function 2:

$\frac {dT} {dr} = 40 \frac {d ((r ^ 2-2r))} {dr}$

$\frac {dT} {dr} = 40 (2r-2)$

Rewriting

$\frac {dT} {dr} = 80r-80$

Answer:

The instantaneous variation of the temperature with respect to r is given by:

Assuming function 1:

$\frac {dT} {dr} = 80r + 80$

Assuming Function 2:

$\frac {dT} {dr} = 80r-80$

6 0

3 years ago

Integrate the following

enot [183]

I suppose you mean to have the entire numerator under the square root?

$\displaystyle\int_2^4\frac{\sqrt{x^2-4}}{x^2}\,\mathrm dx$

We can use a trigonometric substitution to start:

$x=2\sec t\implies\mathrm dx=2\sec t\tan t\,\mathrm dt$

Then for $x=2$ , $t=\sec^{-1}1=0$ ; for $x=4$ , $t=\sec^{-1}2=\frac\pi3$ . So the integral is equivalent to

$\displaystyle\int_0^{\pi/3}\frac{\sqrt{(2\sec t)^2-4}}{(2\sec t)^2}(2\sec t\tan t)\,\mathrm dt=\int_0^{\pi/3}\frac{\tan^2t}{\sec t}\,\mathrm dt$

We can write

$\dfrac{\tan^2t}{\sec t}=\dfrac{\frac{\sin^2t}{\cos^2t}}{\frac1{\cos t}}=\dfrac{\sin^2t}{\cos t}=\dfrac{1-\cos^2t}{\cos t}=\sec t-\cos t$

so the integral becomes

$\displaystyle\int_0^{\pi/3}(\sec t-\cos t)\,\mathrm dt=\boxed{\ln(2+\sqrt3)-\frac{\sqrt3}2}$

7 0

3 years ago

d1i1m1o1n [39]

Answer:

D

Step-by-step explanation:

7 0

4 years ago

Find the missing variable for a parallelogram:<br> A = 25in2<br> h =<br> b = 3.4 in<br> (1in=2.54cm)

Alex777 [14]

$A=Bh$

$h=A/B$

$h=25/3.4 = 7.4 \ in = 18.7 \ cm$

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3 years ago

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Brainliest to whoever gets it right!!!!

IgorC [24]

Answer:

c

Step-by-step explanation:

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4 years ago

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