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Dafna11 [192]
2 years ago
10

A researcher is interested in testing the perceived effort (PE) of runners under three different terrains. PE is measured on a s

cale that ranges from 0 to 15, where higher scores indicate greater effort. The three terrains were a running track (i.e. the control), a mostly flat sidewalk, and a hilly neighborhood. Participants were randomly assigned to only one type of terrain. What type of analysis should the researcher conduct
Mathematics
1 answer:
dsp732 years ago
5 0
I don’t know i’m only in eighth grade
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Evaluate a+b for a=2 and b=3
hoa [83]

Answer:

5

Step-by-step explanation:

a + b

2 + 3

5

8 0
3 years ago
Read 2 more answers
Find \(\int \dfrac{x}{\sqrt{1-x^4}}\) Please, help
ki77a [65]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2867785

_______________


Evaluate the indefinite integral:

\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-x^4}}\,dx}\\\\\\ \mathsf{=\displaystyle\int\! \frac{1}{2}\cdot 2\cdot \frac{1}{\sqrt{1-(x^2)^2}}\,dx}\\\\\\ \mathsf{=\displaystyle \frac{1}{2}\int\! \frac{1}{\sqrt{1-(x^2)^2}}\cdot 2x\,dx\qquad\quad(i)}


Make a trigonometric substitution:

\begin{array}{lcl}
\mathsf{x^2=sin\,t}&\quad\Rightarrow\quad&\mathsf{2x\,dx=cos\,t\,dt}\\\\
&&\mathsf{t=arcsin(x^2)\,,\qquad 0\ \textless \ x\ \textless \ \frac{\pi}{2}}\end{array}


so the integral (i) becomes

\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{1-sin^2\,t}}\cdot cos\,t\,dt\qquad\quad (but~1-sin^2\,t=cos^2\,t)}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{cos^2\,t}}\cdot cos\,t\,dt}

\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{cos\,t}\cdot cos\,t\,dt}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\int\!\f dt}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\,t+C}


Now, substitute back for t = arcsin(x²), and you finally get the result:

\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=\frac{1}{2}\,arcsin(x^2)+C}          ✔

________


You could also make

x² = cos t

and you would get this expression for the integral:

\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=-\,\frac{1}{2}\,arccos(x^2)+C_2}          ✔


which is fine, because those two functions have the same derivative, as the difference between them is a constant:

\mathsf{\dfrac{1}{2}\,arcsin(x^2)-\left(-\dfrac{1}{2}\,arccos(x^2)\right)}\\\\\\
=\mathsf{\dfrac{1}{2}\,arcsin(x^2)+\dfrac{1}{2}\,arccos(x^2)}\\\\\\
=\mathsf{\dfrac{1}{2}\cdot \left[\,arcsin(x^2)+arccos(x^2)\right]}\\\\\\
=\mathsf{\dfrac{1}{2}\cdot \dfrac{\pi}{2}}

\mathsf{=\dfrac{\pi}{4}}         ✔


and that constant does not interfer in the differentiation process, because the derivative of a constant is zero.


I hope this helps. =)

6 0
3 years ago
Large cheese pizzas cost $5 each and large one-topping pizzas cost $6 each.
dsp73

Answer:

T=5c+6d

Step-by-step explanation:

The correct question is

Large cheese pizzas cost $5 each and large one-topping pizzas cost $6 each.

Write an equation that represents the total cost, T, of c large cheese pizzas and d large one-topping pizzas.

Let

T -----> the total cost

c ----> the number of large cheese pizzas

d ---> the number of large one -topping pizzas

we know that

The total cost (T) is equal to the number of large cheese pizzas (c) multiplied by it cost ($5) plus the number of large one -topping pizzas (d) multiplied by it cost ($6)

T=5c+6d

8 0
3 years ago
Two different simple random samples are drawn from two different populations. The first sample consists of 20 people with 9 havi
QveST [7]

Answer: (-0.48,\ -0.04)

Step-by-step explanation:

The confidence interval for the difference of two population proportion is given by :-

(p_1-p_2)\pm z_{\alpha/2}\sqrt{\dfrac{p_1(1-p_1)}{n_1}+\dfrac{p_2(1-p_2)}{n_2}}

Given : The first sample consists of 20 people with 9 having a common attribute.

Here, n_1=20 , p_1=\dfrac{9}{20}=0.45

n_2=2100 , p_1=\dfrac{1492 }{2100}\approx0.71

Significance level : \alpha=1-0.95=0.05

Critical value : z_{\alpha/2}=1.96

A 95% confidence interval for the difference of two population proportion will be :-

(0.45-0.71)\pm z_{\alpha/2}\sqrt{\dfrac{0.45(1-0.45)}{20}+\dfrac{0.71(1-0.71)}{2100}}\\\\\approx -0.26\pm0.22\\\\=(-0.26-0.22,-0.26+0.22)\\\\=(-0.48,\ -0.04)

3 0
4 years ago
Self Check E
goldfiish [28.3K]

Answer:

35 dollars for each lawn . If you mows 32 thats mean you will get 32 × 35 Which is equal to 1120 dollars. So the answear is B

5 0
4 years ago
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