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Ann [662]
3 years ago
5

If IG = 27, find CI.

Mathematics
1 answer:
Alina [70]3 years ago
7 0

Answer: 54

Step-by-step explanation:

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Please help, very urgent!
STALIN [3.7K]

Answer:

0.36

Step-by-step explanation:

Given, a circle with radius, r=4 \ cm

Area of circle =\pi r^2

Substitute r=4

Area of whole circle=\pi r^2\\=\pi\times 4^2\\=16\pi \\=50.2654\ cm^2

Square is inscribed in it whose each side is 4\sqrt{2}\ cm

Area of square =side^2\\=(4\sqrt{2} )^2\\=16\times 2\\=32 \ cm^2

We can see that area of the white circle

= area of the whole circle - area of square

=50.2654-32\\=18.2654 \ cm^2

Probability of falling a random point within the white circle

=\frac{area\ of\ white\ circle}{area\ of\ whole\ circle}

=\frac{18.2654}{50.2654} \\=0.3633

Rounding to nearest hundredth.

Probability of falling a random point within the white circle would be 0.36

5 0
4 years ago
Como resolver função logaritmicas
Alex_Xolod [135]
<span>No sé la respuesta. AMO MUCHA MÚSICA


Good luck! :)</span>
7 0
3 years ago
Simplify Completely<br> 6x^4 + 9x^2 + 12x over 3x
Aleksandr-060686 [28]
6x^4/3x=2x^3 9x^2/3x=3x 12x/3x=4 well final answer is 2x^3+3x+4
3 0
3 years ago
Read 2 more answers
exercise alone. Two methods of collecting data have been proposed. Method I: Recruit volunteers who are willing to participate.
aniked [119]

Answer:

I) If method I is used, population of generalization will include all those people who may have varying exercising habits or routines.  They may or may not have a regular excersing habit.  In his case sample is taken from a more diverse population

II) Population of generalization will include people who will have similar excersing routines and habits if method II is used since sample is choosen from a specific population

Step-by-step explanation:

past excercising habits may affect the change in intensity to a targeted excersise in different manner. So in method I a greater diversity is included and result of excersing with or without a trainer will account for greater number of variables than method II.

5 0
4 years ago
Simplify each expression to a single trig function or number cosx(secx-cosx)
AlladinOne [14]
I did this test  b4, yours is answer #number 12

Convert things to their basic forms. 
<span>Remember a few identities </span>
<span>sin^2 + cos^2 = 1 so </span>
<span>sin^2 = 1 - cos^2 and </span>
<span>cos^2 = 1 - sin^2 </span>

<span>I'm going to skip typing the theta symbol, just to make things faster. Just assume it is there and fill it in as you work the problems. </span>

<span>Follow along to see how each problem was worked out. You'll catch on to the general technique. </span>

<span>====== </span>
<span>1. sec θ sin θ </span>
<span>1/cos * sin = sin/cos = tan </span>

<span>2. cos θ tan θ </span>
<span>cos * sin/cos = sin </span>

<span>3. tan^2 θ- sec^2 θ </span>
<span>sin^2 / cos^2 - 1/cos^2 </span>
<span>(sin^2 - 1)/cos^2 </span>
<span>-(1-sin^2)/cos^2 </span>
<span>-cos^2/cos^2 </span>
<span>-1 </span>

<span>4. 1- cos^2θ </span>
<span>sin^2 </span>

<span>5. (1-cosθ)(1+cosθ) </span>
<span>Remember (a+b)(a--b) = a^2 - b^2 </span>
<span>1-cos^2 = sin^2 </span>

<span>6. (secx-1) (secx+1) </span>
<span>sec^2 -1 </span>
<span>1/cos^2 - 1 </span>
<span>1/cos^2 - cos^2/cos^2 </span>
<span>(1-cos^2)/cos^2 </span>
<span>sin^2/cos62 </span>
<span>tan^2 </span>

<span>7. (1/sin^2A)-(1/tan^2A) </span>
<span>1/sin^2 - 1/(sin^2/cos^2) </span>
<span>1/sin^2 - cos^2/sin^2 </span>
<span>(1-cos^2)/sin^2 </span>
<span>sin^2/sin^2 </span>
<span>1 </span>

<span>8. 1- (sin^2θ/tan^2θ) </span>
<span>1-sin^2/(sin^2/cos^2) </span>
<span>1 - sin^2*cos^2/sin^2 </span>
<span>1-cos^2 </span>
<span>sin^2 </span>


<span>9. (1/cos^2θ)-(1/cot^2θ) </span>
<span>1/cos^2 - 1/(cos^2/sin^2) </span>
<span>1/cos^2 - sin^2/cos^2 </span>
<span>(1-sin^2)/cos^2 </span>
<span>cos^2/cos^2 </span>
<span>1 </span>

<span>10. cosθ (secθ-cosθ) </span>
<span>cos *(1/cos - cos) </span>
<span>1-cos^2 </span>
<span>sin^2 </span>

<span>11. cos^2A (sec^2A-1) </span>
<span>cos^2 * (1/cos^2 - 1) </span>
<span>1 - cos^2 </span>
<span>sin^2 </span>


<span>12. (1-cosx)(1+secx)(cosx) </span>
<span>(1-cos)(1+1/cos)cos </span>
<span>(1-cos)(cos + 1) </span>
<span>-(cos-1)(cos+1) </span>
<span>-(cos^2 - 1) </span>
<span>-(-sin^2) </span>
<span>sin^2 </span>

<span>13. (sinxcosx)/(1-cos^2x) </span>
<span>sin*cos/sin^2 </span>
<span>cos/sin </span>
<span>cot </span>

<span>14. (tan^2θ/secθ+1) +1 </span>
<span>(sin^2/cos^2)/(1/cos) + 2 </span>
<span>sin^2/cos + 2 </span>
<span>sin*tan + 2 </span>
5 0
3 years ago
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