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yaroslaw [1]
3 years ago
10

Using an applet, we found that the theory-based test results in a t-statistic of -7.25 and p-value less than 0.0001. State your

conclusion in the context of the study.
a. We have very strong evidence that the long-run average guess of the population size of Milwaukee, Wisconsin is smaller with the anchor of Chicago than with the anchor of Green Bay.
b. We have very strong evidence that the long-run average guess of the population size of Milwaukee, Wisconsin is smaller with the anchor of Green Bay than with the anchor of Chicago.
c. We have very weak evidence that the long-run average guess of the population size of Milwaukee, Wisconsin is larger with the anchor of Green Bay than with the anchor of Chicago.
d. We have very weak evidence that the long-run average guess of the population size of Milwaukee, Wisconsin is larger with the anchor of Chicago than with the anchor of Green Bay.
Mathematics
1 answer:
11111nata11111 [884]3 years ago
7 0

Answer:

b. We have very strong evidence that the long-run average guess of the population size of Milwaukee, Wisconsin is smaller with the anchor of Green Bay than with the anchor of Chicago.

Step-by-step explanation:

Smaller p-value indicates a strong evidence in favor of the alternative hypothesis which means we can reject the null hypothesis. In the given scenario the p-value is 0.0001 which is very small. There is sufficient evidence for the rejection of null hypothesis.

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Determine whether the two expressions are equivalent. If so, tell what property is applied. If not, explain why. 10 ÷ 5 and 5 ÷
nikklg [1K]

Answer:

no, its not equivalent

Step-by-step explanation:

10/5= 2

5/10=0.5

5 0
3 years ago
Order these numbers from greatest to least: -4 , 1/4 , 0 , 4 , -3 1/2 , 7/4 , 5/4
Nadya [2.5K]

Answer:-4, -3 1/2, 0, 1/4, 5/4, 7/4, 4

Step-by-step explanation: please give brainliest and thankyou

6 0
3 years ago
Dy/dx = 2xy^2 and y(-1) = 2 find y(2)
Anarel [89]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2887301

—————

Solve the initial value problem:

   dy
———  =  2xy²,      y = 2,  when x = – 1.
   dx


Separate the variables in the equation above:

\mathsf{\dfrac{dy}{y^2}=2x\,dx}\\\\
\mathsf{y^{-2}\,dy=2x\,dx}


Integrate both sides:

\mathsf{\displaystyle\int\!y^{-2}\,dy=\int\!2x\,dx}\\\\\\
\mathsf{\dfrac{y^{-2+1}}{-2+1}=2\cdot \dfrac{x^{1+1}}{1+1}+C_1}\\\\\\
\mathsf{\dfrac{y^{-1}}{-1}=\diagup\hspace{-7}2\cdot \dfrac{x^2}{\diagup\hspace{-7}2}+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{y}=x^2+C_1}

\mathsf{\dfrac{1}{y}=-(x^2+C_1)}


Take the reciprocal of both sides, and then you have

\mathsf{y=-\,\dfrac{1}{x^2+C_1}\qquad\qquad where~C_1~is~a~constant\qquad (i)}


In order to find the value of  C₁  , just plug in the equation above those known values for  x  and  y, then solve it for  C₁:

y = 2,  when  x = – 1. So,

\mathsf{2=-\,\dfrac{1}{1^2+C_1}}\\\\\\
\mathsf{2=-\,\dfrac{1}{1+C_1}}\\\\\\
\mathsf{-\,\dfrac{1}{2}=1+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-1=C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-\dfrac{2}{2}=C_1}

\mathsf{C_1=-\,\dfrac{3}{2}}


Substitute that for  C₁  into (i), and you have

\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}}\\\\\\
\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}\cdot \dfrac{2}{2}}\\\\\\
\mathsf{y=-\,\dfrac{2}{2x^2-3}}


So  y(– 2)  is

\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot (-2)^2-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot 4-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{8-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{5}}\quad\longleftarrow\quad\textsf{this is the answer.}


I hope this helps. =)


Tags:  <em>ordinary differential equation ode integration separable variables initial value problem differential integral calculus</em>

7 0
3 years ago
Solve 2x + 4 &gt; 16<br><br>A. x &gt; 10<br><br>B. x &gt; 6<br><br>C. n &lt; 6<br><br>D. x &lt; 10​
Alexandra [31]

Answer:

The answer is B ( X>6 )

Step-by-step explanation:

2x+4= 16

2x = 16 - 4

2x = 12

x = 12/2

x= 6

x>6

8 0
3 years ago
Consider these functions:
valkas [14]
I think the answer might be B
6 0
2 years ago
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