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Firdavs [7]
3 years ago
14

In circle O, central angle AOB has a measure of 90°. Which of the following is not true? segment OA is a radius segments OA and

OB are perpendicular arc AB is a semicircle

Mathematics
2 answers:
Zinaida [17]3 years ago
5 0

Answer:

Arc AB is a semicircle, Not true.

Step-by-step explanation:

In circle O, central angle AOB has measure of 90°

We need to choose statement which is not true. We can see the attachment to shape of circle.  

As we can see O is the center of circle. A and B are two points on circle.  

So, OA and OB must be radius of this circle.  

1) Segment OA is a radius ( TRUE )

2) Segment OA and OB are perpendicular (TRUE)

Because central angle makes 90°  

3) arc AB is a semicircle (FALSE)

Because AOB is not 180°


lesantik [10]3 years ago
3 0
The last one: the arc AB is a semicircle. 

a semicircle is 180 degrees
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as 205 is the term greater than the 1st term of AP and the difference is in decreasing order

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D

Step-by-step explanation:

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2 sin² x-sin x=0

sin x(2 sin x-1)=0

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2 sin x-1=0

2 sin x=1

sin x=1/2= sin π/6,sin (π-π/6)

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4 0
3 years ago
Given that (-8,-5) is on the graph of f(x), find the corresponding point for the function f(x-2)
Volgvan
This one works the same way the last one did.

If -8 is x-2, then x = -6. The corresponding point is (-6, -5).
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Which expression is equivalent to -3(4x - 2)- 2x
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Step-by-step explanation:

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3 years ago
In a class of 19 students, 3 are math majors. A group of four students is chosen at random. (Round your answers to four decimal
KatRina [158]

Answer:

(a) The probability is 0.4696

(b) The probability is 0.5304

(c) The probability is 0.0929

Step-by-step explanation:

The total number of ways in which we can select k elements from a group n elements is calculate as:

nCx=\frac{n!}{x!(n-x)!}

So, the number of ways in which we can select four students from a group of 19 students is:

19C4=\frac{19!}{4!(19-4)!}=3,876

On the other hand, the number of ways in which we can select four students with no math majors is:

(16C4)*(3C0)=(\frac{16!}{4!(16-4)!})*(\frac{3!}{0!(3-0)!})=1820

Because, we are going to select 4 students form the 16 students that aren't math majors and select 0 students from the 3 students that are majors.

At the same way, the number of ways in which we can select four students with one, two and three math majors are 1680, 360 and 16 respectively, and they are calculated as:

(16C3)*(3C1)=(\frac{16!}{3!(16-3)!})*(\frac{3!}{1!(3-1)!})=1680

(16C2)*(3C2)=(\frac{16!}{2!(16-2)!})*(\frac{3!}{2!(3-1)!})=360

(16C1)*(3C3)=(\frac{16!}{1!(16-1)!})*(\frac{3!}{3!(3-3)!})=16

Then, the probability that the group has no math majors is:

P=\frac{1820}{3876} =0.4696

The probability that the group has at least one math major is:

P=\frac{1680+360+16}{3876} =0.5304

The probability that the group has exactly two math majors is:

P=\frac{360}{3876} =0.0929

6 0
3 years ago
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