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Mariulka [41]
3 years ago
8

Choose the expressions that are equivalent to m∠AOB. Select all that apply.

Mathematics
2 answers:
Tanzania [10]3 years ago
6 0

Answer:

A, B, D, G, and H should be correct

Step-by-step explanation:

I've done the same problem and those were the answers shown

Nata [24]3 years ago
5 0

Answer:

A, B, D, G, and H

Explanation:

i just did it :)

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What is this pls help
Julli [10]

Answer:

5z

Step-by-step explanation:

3z+5z-3z(3z gets cancelled)

5z

3 0
3 years ago
Read 2 more answers
BRAINLIEST TO WHOEVER GETS IT RIGHT!! Kyle has to run 3 1/2 laps for his workout. He is 1/2 finished. How many laps has Kyle ran
8090 [49]

Answer:

A. 1 /14

Step-by-step explanation:

3 1/2 / 1/2 = 1 /14

3 0
3 years ago
Pleaseeee help me i have very little time
VashaNatasha [74]

Answer:

212 feet

Step-by-step explanation:

Please let me know if you want me to add an explanation as to why this is the answer. I can definitely do that, I just wouldn’t want to write it if you don’t want me to :)

7 0
3 years ago
Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen
k0ka [10]

Let a,b,c be the randomly selected lengths. Without loss of generality, suppose a[tex]P(A + B \ge C) = P(A + B - C \ge 0)

where A,B,C are independent random variables with the same uniform distribution on [0, 1].

By their mutual independence, we have

P(A=a,B=b,C=c) = P(A=a) \times P(B=b) \times P(C=c)

so that the joint density function is

P(A=a,B=b,C=c) = \begin{cases}1 & \text{if }(a,b,c)\in[0,1]^3 \\ 0 & \text{otherwise}\end{cases}

where [0,1]^3=[0,1]\times[0,1]\times[0,1] is the cube with vertices at (0, 0, 0) and (1, 1, 1).

Consider the plane

a + b - c = 0

with (a,b,c)\in\Bbb R^3. This plane passes through (0, 0, 0), (1, 0, 1), and (0, 1, 1), and thus splits up the cube into one tetrahedral region above the plane and the rest of the cube under it. (see attached plot)

The point (0, 0, 1) (the vertex of the cube above the plane) does not belong the region a+b-c\ge0, since 0+0-1=-1. So the probability we want is the volume of the bottom "half" of the cube. We could integrate the joint density over this set, but integrating over the complement is simpler since it's a tetrahedron.

Then we have

\displaystyle P(A+B-C\ge0) = 1 - P(A+B-C < 0) \\\\ ~~~~~~~~ = 1 - \int_0^1\int_0^{1-a}\int_{a+b}^1 P(A=a,B=b,C=c) \, dc\,db\,da \\\\ ~~~~~~~~ = 1 - \int_0^1 \int_0^{1-a} (1 - a - b) \, db \, da \\\\ ~~~~~~~~ = 1 - \int_0^1 \frac{(1-a)^2}2\,da \\\\ ~~~~~~~~ = 1 - \frac16 = \boxed{\frac56}

5 0
2 years ago
Simplify 5/8x -2y +3/4x-8y
AleksandrR [38]

Answer:

On simplification,  \frac{5}{8}x - 2y + \frac{3}{4}x -8y   =   (\frac{11}{8}) x - 10y

Step-by-step explanation:

Here, the given expression is:

\frac{5}{8}x - 2y + \frac{3}{4}x -8y

Now, we can perform operations only on LIKE TERMS,

So, in this expression, separate the like terms we get:

\frac{5}{8}x - 2y + \frac{3}{4}x -8y  =  \frac{5}{8}x +  \frac{3}{4}x - 2y -8y\\= (\frac{5}{8} +  \frac{3}{4})x -(2y + 8y)  = (\frac{5 + 3(2)}{8}) x - (10y)\\= (\frac{11}{8}) x - 10y\\\implies  \frac{5}{8}x - 2y + \frac{3}{4}x -8y   =   (\frac{11}{8}) x - 10y

Hence, on simplification,  \frac{5}{8}x - 2y + \frac{3}{4}x -8y   =   (\frac{11}{8}) x - 10y

8 0
3 years ago
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