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2 years ago

Which of the following appear in the diagram below? Check all that apply. A. vec CE B. angle DCE OC . overline AB OD. angle CDE

Mathematics

1 answer:

Ira Lisetskai [31]2 years ago

7 0

Answer:

OPTION A

OPTION B

OPTION C

Step-by-step explanation:

By definition a "Ray" is a part of a line that has one endpoint (or point of origin) and continues in one direction forever. Knowing this, you can analize the diagram provided:

- You can observe in the given diagram that there are three rays. These are:

$AB, CD$ and $CE$

- You can notice that an angle is formed by two rays ( Ray $CD$ and ray $CE$ ) with the same endpoint (Point C).

Therefore, the angle formed is:

$\angle DCE$

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In the following diagram \overline{DE} \parallel \overline{FG}

Gnesinka [82]

Answer:

<x = 31°

Step-by-step explanation:

m<BCA = m<GCJ (vertical angles)

m<BCA = 59° (substitution)

Since line KL is perpendicular to line FG, the angle formed at point B is 90°.

Therefore, m<ABC = 90°

m<BAC + m<ABC + m<BCA = 180° (sum of triangle)

m<BAC + 90° + 59° = 180° (Substitution)

m<BAC + 149° = 180°

m<BAC = 180° - 149°

m<BAC = 31°

<x = <BAC (vertical angles)

m<x = 31° (substitution)

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2 years ago

June needs 48 gallons of punch

LenaWriter [7]

Answer:

David needs 36.style="color:rgb(72,72,72);background-color:transparent;"> gallons of punch for a party and has two different coolers to carry it in. The bigger cooler is 5. times as large as the smaller cooler.

8 0

2 years ago

Read 2 more answers

Discuss the validity of the following statement. If the statement is always true, explain why. If not, give a counterexample.

Zarrin [17]

Correction:

Because F is not present in the statement, instead of working onP(E)P(F) = P(E∩F), I worked on

P(E∩E') = P(E)P(E').

Answer:

The case is not always true.

Step-by-step explanation:

Given that the odds for E equals the odds against E', then it is correct to say that the E and E' do not intersect.

And for any two mutually exclusive events, E and E',

P(E∩E') = 0

Suppose P(E) is not equal to zero, and P(E') is not equal to zero, then

P(E)P(E') cannot be equal to zero.

P(E)P(E') ≠ 0

This makes P(E∩E') different from P(E)P(E')

Therefore,

P(E∩E') ≠ P(E)P(E') in this case.

8 0

2 years ago

Two students from a group of eight boys and 12 girls are sent to represent the school in a parade. If the students are chosen at

kari74 [83]

Answer:

Step-by-step explanation:

* Lets explain how to find the probability of an event

- The probability of an Event = Number of favorable outcomes ÷ Total

number of possible outcomes

- P(A) = n(E) ÷ n(S) , where

# P(A) means finding the probability of an event A

# n(E) means the number of favorable outcomes of an event

# n(S) means set of all possible outcomes of an event

- Probability of event not happened = 1 - P(A)

- P(A and B) = P(A) . P(B)

* Lets solve the problem

- There is a group of students

- There are 8 boys and 12 girls in the group

∴ There are 8 + 12 = 20 students in the group

- The students are sent to represent the school in a parade

- Two students are chosen at random

∴ P(S) = 20

- The students that chosen are not both girls

∴ The probability of not girls = 1 - P(girls)

∵ The were 20 students in the group

∵ The number of girls in the group was 12

∴ The probability of chosen a first girl = 12/20

∵ One girl was chosen, then the number of girls for the second

choice is less by 1 and the total also less by 1

∴ The were 19 students in the group

∵ The number of girls in the group was 11

∴ The probability of chosen a second girl = 11/19

- The probability of both girls is P(1st girle) . P(2nd girl)

∴ The probability of both girls = (12/20) × (11/19) = 33/95

- To find the probability of both not girls is 1 - P(both girls)

∴ P(not both girls) = 1 - (33/95) = 62/95

* The probability that the students chosen are not both girls is 62/95

3 0

2 years ago

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How would you choose to reduce the system shown to a 2 × 2? Explain why you would choose this approach. –3x + y – 2z = 10 (1) 5x

gladu [14]

-3x + y - 2z = 10 |* -1
3x - y +2z = -10
5x -2y -2z = 12
--------------------------- I add these equations term by term
8x - 3y = 2

-3x + y - 2z =10 ⇒ -3x + y - 2z =10
x -y +z = 23 | *2 2x - 2y + 2z = 46
----------------------------- I add these eq.
-x -y = 56

8x - 3y = 2
-x -y = 56

this is the system after i reduce it ( it has only two variables x and y)

4 0

2 years ago

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