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Alex787 [66]
2 years ago
13

I WILL GIVE BRAINLIEST TO THE FIRST PERSON!A spinner has 8 equal-sized sections. The probability that the spinner will stop abov

e a blue section is 3/4. Which best describes the probability of the spinner stopping above a blue section? A. impossible B. unlikely C. likely D. certain
Mathematics
2 answers:
zloy xaker [14]2 years ago
6 0
C. Likely. I hope this helped. Also it's 3/4 instead of 4/4 , if it was 4/4 it would be certain but 3/4 would be likely because it's a 1/4 chance it could land on another color
FromTheMoon [43]2 years ago
3 0
The chances of it spinning on blue is likely

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The apartment building where James lives was built with bricks that total 41 10000 write this number in standard form
Allisa [31]
Do you mean millions?  If so, please write 4,110,000, which is in terms of millions and is in standard form.  Otherwise, ensure that you have copied down this problem correctly.
8 0
3 years ago
What are the quotient and remainder of (3x^4+ 2x^2 - 6x + 1) /(x + 1)
quester [9]

Answer:

all work is shown and pictured

4 0
2 years ago
Radio signals travel at a rate of 3 x 10^8 meters per second. How many seconds would it take for a radio signal to travel from a
wel
Time = Distance / Speed   [ Already mentioned ]

Then, substitute the known values, 

t = (9.6 × 10⁶) / (3 × 10⁸)

t = 3.2 × 10⁻² s

In short, Your Answer would be Option B

Hope this helps!
4 0
3 years ago
Given the list of terms 1/3​, 1, 5/3, 7/3,.... Find the 15th and −16th term.
qwelly [4]

Answer:

15th term =29/3

16th term = 31/3

Step-by-step explanation:

Given an arithmetic sequence with the first term a1 and the common difference d , the nth (or general) term is given by an=a1+(n−1)d .

First we find the 15th term

n=15

a1=1/3

d=1 - 1/3 = 2/3

Solution

1/3+(15-1)2/3

1/3+28/3

(1+28)/3

29/3

Lets find the 16th term

1/3+(16-1)2/3

1/3+30/3

(1+30)/3

31/3

8 0
3 years ago
A torus is formed by rotating a circle of radius r about a line in the plane of the circle that is a distance R (> r) from th
jeyben [28]

Consider a circle with radius r centered at some point (R+r,0) on the x-axis. This circle has equation

(x-(R+r))^2+y^2=r^2

Revolve the region bounded by this circle across the y-axis to get a torus. Using the shell method, the volume of the resulting torus is

\displaystyle2\pi\int_R^{R+2r}2xy\,\mathrm dx

where 2y=\sqrt{r^2-(x-(R+r))^2}-(-\sqrt{r^2-(x-(R+r))^2})=2\sqrt{r^2-(x-(R+r))^2}.

So the volume is

\displaystyle4\pi\int_R^{R+2r}x\sqrt{r^2-(x-(R+r))^2}\,\mathrm dx

Substitute

x-(R+r)=r\sin t\implies\mathrm dx=r\cos t\,\mathrm dt

and the integral becomes

\displaystyle4\pi r^2\int_{-\pi/2}^{\pi/2}(R+r+r\sin t)\cos^2t\,\mathrm dt

Notice that \sin t\cos^2t is an odd function, so the integral over \left[-\frac\pi2,\frac\pi2\right] is 0. This leaves us with

\displaystyle4\pi r^2(R+r)\int_{-\pi/2}^{\pi/2}\cos^2t\,\mathrm dt

Write

\cos^2t=\dfrac{1+\cos(2t)}2

so the volume is

\displaystyle2\pi r^2(R+r)\int_{-\pi/2}^{\pi/2}(1+\cos(2t))\,\mathrm dt=\boxed{2\pi^2r^2(R+r)}

6 0
2 years ago
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