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3 years ago

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olganol [36]3 years ago

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Prove the segments joining the midpoint of consecutive sides of an isosceles trapezoid form a rhombus. use the distance formula

riadik2000 [5.3K]

In order to prove this, we have to put the trapezoid to the coordinate system. In the attached photo you can see how it has to be put. The coordinates for the vertices of trapezoid written according to the midpoint principle. By using the distance between two points formula, we can find the coordinates for the vertices of the rhombus.
$D_{x} = \frac{-2b-2a}{2}=-b-a$ and $D_{y}= \frac{2c+0}{2}=c$ . The coordinates of D is $(-b-a, c)$
$E_{x} = \frac{-2b+2b}{2}=0$ and $E_{y}= \frac{2c+2c}{2}=2c$ . The coordinates of E is $(0,2c)$
Since we have the reflection in this graph, the coordinates of F is $(b+a, c)$
And the coordinates of G is (0,0).

Using the distance formula, we can find that
$DE= \sqrt{(-b-a)^{2}+ c^{2}}=\sqrt{(b+a)^{2} + c^{2}}$
$EF= \sqrt{(b+a)^{2} + c^{2} }$
$GF= \sqrt{(b+a )^{2} + c^{2} }$
$DG= \sqrt{(b+a)^{2}+ c^{2}$

Since all the sides are equal this completes our proof. Additionally, we can find the distances of EG and DF in order to show that the diagonals of this rhombus are not equal. So that it is not a square, but rhombus.

6 0

3 years ago

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Which choice provides the best evidence for the

vodka [1.7K]

Correct option is B, Lines 39-42 ("Normally . . . community")

The "matter of urgency," as Akira explains in lines 39–42, is that he has "an opportunity to go to America, as dentist for Seattle's Japanese community." Chie's response to Akira's marriage proposal is necessary for him to determine whether or not to take the position in Seattle.

The answers to the previous question are not well supported by the evidence in Choices A, C, and D. Choice A is wrong since Akira apologizes for disturbing Chie's peaceful evening in line 39. Lines 58–59 discuss the gravity of Akira's demand, not its immediacy, hence option C is incorrect.

Line 73 merely demonstrates that Akira's suggestion "startled" Chie, not why his request is time-sensitive, hence option D is incorrect.

Here's another question with an answer similar to this Chie and Akira:

brainly.com/question/21336173?

#SPJ4

6 0

2 years ago

Help calculus module 8 DBQ<br><br> please show work

igor_vitrenko [27]

1. The four subintervals are [0, 2], [2, 3], [3, 7], and [7, 8]. We construct trapezoids with "heights" equal to the lengths of each subinterval - 2, 1, 4, and 1, respectively - and the average of the corresponding "bases" equal to the average of the values of $R(t)$ at the endpoints of each subinterval. The sum is then

$\dfrac{R(0)+R(2)}2(2-0)+\dfrac{R(2)+R(3)}2(3-2)+\dfrac{R(3)+R(7)}2(7-3)+\dfrac{R(7)+R(8)}2(7-8)=\boxed{24.83}$

which is measured in units of gallons, hence representing the amount of water that flows into the tank.

2. Since $R$ is differentiable, the mean value theorem holds on any subinterval of its domain. Then for any interval $[a,b]$ , it guarantees the existence of some $c\in(a,b)$ such that

$\dfrac{R(b)-R(a)}{b-a)=R'(c)$

Computing the difference quotient over each subinterval above gives values of 0.275, 0.3, 0.3, and 0.26. But just because these values are non-zero doesn't guarantee that there is definitely no such $c$ for which $R'(c)=0$ . I would chalk this up to not having enough information.

3. $R(t)$ gives the rate of water flow, and $R(t)\approx W(t)$ , so that the average rate of water flow over [0, 8] is the average value of $W(t)$ , given by the integral

$R_{\rm avg}=\displaystyle\frac1{8-0}\int_0^8\ln(t^2+7)\,\mathrm dt$

If doing this by hand, you can integrate by parts, setting

$u=\ln(t^2+7)\implies\mathrm du=\dfrac{2t}{t^2+7}\,\mathrm dt$

$\mathrm dv=\mathrm dt\implies v=t$

$R_{\rm avg}=\displaystyle\frac18\left(t\ln(t^2+7)\bigg|_{t=0}^{t=8}-\int_0^8\frac{2t^2}{t^2+7}\,\mathrm dt\right)$

For the remaining integral, consider the trigonometric substitution $t=\sqrt 7\tan s$ , so that $\mathrm dt=\sqrt 7\sec^2s\,\mathrm ds$ . Then

$R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}\frac{7\tan^2s}{7\tan^2s+7}\sec^2s\,\mathrm ds$

$R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}\tan^2s\,\mathrm ds$

$R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}(\sec^2s-1)\,\mathrm ds$

$R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\left(\tan s-s\right)\bigg|_{s=0}^{s=\tan^{-1}(8/\sqrt7)}$

$R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\left(\tan\left(\tan^{-1}\frac8{\sqrt7}\right)-\tan^{-1}\frac8{\sqrt7}\right)$

$\boxed{R_{\rm avg}=\displaystyle\ln71-2+\frac{\sqrt7}4\tan^{-1}\frac8{\sqrt7}}$

or approximately 3.0904, measured in gallons per hour (because this is the average value of $R$ ).

4. By the fundamental theorem of calculus,

$g'(x)=f(x)$

and $g(x)$ is increasing whenever $g'(x)=f(x)>0$ . This happens over the interval (-2, 3), since $f(x)=3$ on [-2, 0), and $-x+3>0$ on [0, 3).

5. First, by additivity of the definite integral,

$\displaystyle\int_{-2}^xf(t)\,\mathrm dt=\int_{-2}^0f(t)\,\mathrm dt+\int_0^xf(t)\,\mathrm dt$

Over the interval [-2, 0), we have $f(x)=3$ , and over the interval [0, 6], $f(x)=-x+3$ . So the integral above is

$\displaystyle\int_{-2}^03\,\mathrm dt+\int_0^x(-t+3)\,\mathrm dt=3t\bigg|_{t=-2}^{t=0}+\left(-\dfrac{t^2}2+3t\right)\bigg|_{t=0}^{t=x}=\boxed{6+3x-\dfrac{x^2}2}$

6 0

3 years ago

Out of 120 high school boys and 80 high school girls, 45 tried out for track. Of those 45, 12 were girls.

zhannawk [14.2K]

<span>So there are 33 boys who tried out for track. And this is 27.5 % of the total boys. And there 15% of the girls who tried out for the track. And in total 22.5 % of the total population tried out for track</span>

5 0

3 years ago

Write equivalent fractions for 11/12 and 9/10 using the least common denominator

qwelly [4]

Answer:

55/60 and 54/60

Step-by-step explanation:

The LCD is 60. Multiply numerators and denominators. Hope this helps.

8 0

3 years ago

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