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pantera1 [17]
2 years ago
15

What even is this? Did my teacher have a typo? 56\pi +67i\leq +23

Mathematics
1 answer:
MakcuM [25]2 years ago
6 0

Answer: it's not a typo its a formula

hope this helps

brainliest?

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Midpoint of the segment whose endpoints are (6,9) and (-4,16)
pychu [463]

Answer:

(4,5)

Step-by-step explanation:

(xa+xb/2,ya+yb/2)

(0+8/2,2+8/2)

(8/2,10/2)

(4,5)

hope this helps

3 0
3 years ago
If there are a hundred centimeters in a meter how many centimeters are in 1/2 meters
Darina [25.2K]
50 cm 100 ÷2 =50 1m÷2=1/2m
6 0
3 years ago
Read 2 more answers
Which expression represents the expression 20 x − 10 20 ⁢ x - 10 after it has been factored completely? A. 2 ( 10 x − 5 ) 2 ⁢ (
son4ous [18]

Answer:

A. 2 ( 10 x − 5 )

D. 10 ( 2 x − 1 )

Step-by-step explanation:

Given expression:

20x − 10

Find the equivalent expression

Check all that applies

A. 2 ( 10 x − 5 )

= 20x - 10

B. 2 ( 2 x − 1 )

= 4x - 2

C. 10 ( 20 x − 10 )

= 200x - 100

D. 10 ( 2 x − 1 )

20x - 10

The correct options are

A. 2 ( 10 x − 5 )

D. 10 ( 2 x − 1 )

7 0
3 years ago
PLZ HELP<br> In ⊙C, DE ≅ FG, Which statement must be true? Select all that apply.
Alex Ar [27]

Answer:

E because is covered all angle

6 0
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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