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Rom4ik [11]
1 year ago
7

Given that cos(theta) = -2 radical 5/5, and is in Quadrant II, what is cos(2theta)?

Mathematics
1 answer:
gtnhenbr [62]1 year ago
5 0

cos \theta =  \frac{ - 2 \sqrt{5} }{5}

cos 2\theta = 2cos {}^{2}  \theta - 1 \\ cos2 \theta = 2( \frac{ - 2 \sqrt{5} }{5} ) {}^{2}  - 1 \\ cos2 \theta = 2( \frac{20}{ \\ 25} ) - 1 \\ cos2 \theta =  \frac{40 - 25}{25}  =  \frac{15}{25}  =  \frac{3}{5}

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The histogram will be a repeat of the individual numbers

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3 years ago
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Quick and easy math questions
maw [93]

Answer:

32/68

Step-by-step explanation:

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Determine whether the normal sampling distribution can be used. The claim is p < 0.015 and the sample size is n
Svetradugi [14.3K]

Complete Question

Determine whether the normal sampling distribution can be used. The claim is p < 0.015 and the sample size is n=150

Answer:

Normal sampling distribution can not be used

Step-by-step explanation:

From the question we are told that

    The  null hypothesis is  H_o  :  p =  0.015

     The  alternative hypothesis is    H_a  :  p  <  0.015

     

The  sample size is  n= 150

Generally in order to use  normal sampling distribution  

     The value  np  \ge  5

So  

         np =  0.015 * 150

         np =  2.25

Given that  np < 5   normal sampling distribution  can not be used

8 0
3 years ago
A piece of copper tubing is in the shape of a hexagonal prism. The area of the base of the prism is 3 square centimeters. The vo
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Answer: 17 cm

Step-by-step explanation:

For a regular figure where the area of the base is B, and the height is H, the volume is calculated as:

V = B*H

Here we know that the base of the prism is B = 3 cm^2, and the volume of the prism is 51 cm^3

Then:

B = 3cm^2

V = 51cm^3

If we replace those in the equation above, we get:

51 cm^3 = (3cm^2)*H

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3 years ago
The longest side of the right triangle is 2 feet more than the middle side and the middle side is 1 foot less than twice the sho
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A right triangle's longest side is the hypotenuse
let x=longest, y=middle, and z=shortest
x=y+2
y=2z-1
therefore x=(2z-1)+2=2z+1
find z
z^2+y^2=x^2 by Pythagorean theorem
plug in x and y in terms of z
z^2+(2z-1)^2=(2z+1)^2
z^2+4z^2-4z+1=4z^2+4z+1
subtract the right-hand side's value from the left-hand side's
z^2-8z=0
z(z-8)=0
z=0, 8
z cannot be zero as the sides must have some value to it.
Therefore the shortest side is equal to 8
3 0
3 years ago
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