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Sergeeva-Olga [200]
2 years ago
9

Experiments on learning in animals sometimes measure how long it takes mice to find their way through a maze. Only half of all m

ice complete one particular maze in less than 18 seconds. A research thinks that a loud noise will cause the mice to complete the maze faster. She measures the proportion of 40 mice that completed the maze in less than 18 seconds with noise as a stimulus. The proportion of mice that completed the maze in less than 18 seconds is . The hypotheses for a test to answer the researcher's question are
a) H0: p = 0.5, Ha: p ≠ 0.5.
b) H0: p = 0.5, Ha: p < 0.5.
c) H0: p = 0.5, Ha: p > 0.5.
Mathematics
1 answer:
nlexa [21]2 years ago
8 0

Answer:

Option c (H0: p = 0.5, Ha: p > 0.5) is the correct option.

Step-by-step explanation:

According to the question,

p = 0.5

  • We must check whether in less than 18 seconds the percentage including all razer mice finish a certain labyrinth is above half, which would be p > 0.5. It's over 1/2 the mouse with much less than 18 seconds to finish the maze.
  • Throughout null H0, humans assert that the percentage of the particular period seems to be equivalent to that specified as well as that beneath theoretical frameworks, this same percentage is always < (less than) or > (greater than) or doesn't equal to the category as per the hypothesis matter.

The other given alternatives aren't related to the given explanation or the conditions. So the above is the right option.

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Help me with trigonometry
poizon [28]

Answer:

See below

Step-by-step explanation:

It has something to do with the<em> </em><u><em>Weierstrass substitution</em></u>, where we have

$\int\, f(\sin(x), \cos(x))dx = \int\, \dfrac{2}{1+t^2}f\left(\dfrac{2t}{1+t^2}, \dfrac{1-t^2}{1+t^2} \right)dt$

First, consider the double angle formula for tangent:

\tan(2x)= \dfrac{2\tan(x)}{1-\tan^2(x)}

Therefore,

\tan\left(2 \cdot\dfrac{x}{2}\right)= \dfrac{2\tan(x/2)}{1-\tan^2(x/2)} = \tan(x)=\dfrac{2t}{1-t^2}

Once the double angle identity for sine is

\sin(2x)= \dfrac{2\tan(x)}{1+\tan^2(x)}

we know \sin(x)=\dfrac{2t}{1+t^2}, but sure,  we can derive this formula considering the double angle identity

\sin(x)= 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)

Recall

\sin \arctan t = \dfrac{t}{\sqrt{1 + t^2}} \text{ and } \cos \arctan t = \dfrac{1}{\sqrt{1 + t^2}}

Thus,

\sin(x)= 2 \left(\dfrac{t}{\sqrt{1 + t^2}}\right) \left(\dfrac{1}{\sqrt{1 + t^2}}\right) = \dfrac{2t}{1 + t^2}

Similarly for cosine, consider the double angle identity

Thus,

\cos(x)=  \left(\dfrac{1}{\sqrt{1 + t^2}}\right)^2- \left(\dfrac{t}{\sqrt{1 + t^2}}\right)^2 = \dfrac{1}{t^2+1}-\dfrac{t^2}{t^2+1} =\dfrac{1-t^2}{1+t^2}

Hence, we showed \sin(x) \text { and } \cos(x)

======================================================

5\cos(x) =12\sin(x) +3, x \in [0, 2\pi ]

Solving

5\,\overbrace{\frac{1-t^2}{1+t^2}}^{\cos(x)} = 12\,\overbrace{\frac{2t}{1+t^2}}^{\sin(x)}+3

\implies \dfrac{5-5t^2}{1+t^2}= \dfrac{24t}{1+t^2}+3 \implies  \dfrac{5-5t^2 -24t}{1+t^2}= 3

\implies 5-5t^2-24t=3\left(1+t^2\right) \implies -8t^2-24t+2=0

t = \dfrac{-(-24)\pm \sqrt{(-24)^2-4(-8)\cdot 2}}{2(-8)} = \dfrac{24\pm 8\sqrt{10}}{-16} =  \dfrac{3\pm \sqrt{10}}{-2}

t=-\dfrac{3+\sqrt{10}}{2}\\t=\dfrac{\sqrt{10}-3}{2}

Just note that

\tan\left(\dfrac{x}{2}\right) =  \dfrac{3\pm 8\sqrt{10}}{-2}

and  \tan\left(\dfrac{x}{2}\right) is not defined for x=k\pi , k\in\mathbb{Z}

6 0
3 years ago
→
Rudiy27
It is 9 degrees because at 6am the temperature was -12 degrees and at 10am the temperature was -3 degrees.
7 0
3 years ago
Please help me out I’m struggling
djyliett [7]
I believe the correct answer to your problem is C , you’re welcome
4 0
3 years ago
Read 2 more answers
Han's house is 450 meters from school. Lin’s house is 135 meters closer to school. How far is Lin’s house from school?
just olya [345]
The answe is the answer is 8
8 0
3 years ago
HELPPPP What is the range of g
dimulka [17.4K]

The range is how long the graph extends vertically. So, the lowest value is -9 (since the graph extends down until -9) and the highest value is 9 (since the graph extends up until 9). The lowest value goes in the first box and the highest value goes in the second box. The range is:

-9 ≤ <em>g</em>(<em>t </em>) ≤9


3 0
2 years ago
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