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natita [175]
3 years ago
15

(100 points if answered) (attatched screenshot) Algebra II need question 1 and 2

Mathematics
1 answer:
Elanso [62]3 years ago
3 0
Domain: (-♾,2)U(2,♾), {x | x ≠2}
Range: (-♾,3) U(3,♾), {y | y ≠ 3}

Sorry that’s all I can help with


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RoseWind [281]
-4 - 1 = -5 can be an equation that equals -5.
8 0
3 years ago
Read 2 more answers
Can the a triangle with the side measurement of 3,4,and 6 inches be formed
abruzzese [7]

Answer:

hola a la par y me dijo dígame qué hora prefieres que te hace falta un poco pero no me e sentido así

Step-by-step explanation:

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3 0
3 years ago
PLEASE HELP!!!!<br>how to solve 9x^3-16x^2=0 by factoring<br> WILL GIVE BRAINLIEST IF CORRECT!!!!
andreev551 [17]

Answer:

x=5.3

Step-by-step explanation:

Graph each side of the equation. The solution is the x-value of the point of intersection.

x

≈

5.3

Hope this helps, can you now give me brainliest??

7 0
3 years ago
What is the probability that both of Eduardo's partners for the group project will be girls? StartFraction 9 Over 65 EndFraction
tester [92]

Answer:

StartFraction 24 Over 65 EndFraction

Step-by-step explanation:

Total number of students = 26

Number of boys = 10

Number of girls = 26-10

=16

Eduardo has to pull two names out of the hat without replacing them.

First name

Probability= Favourable outcome/Total outcome

Probability of girls=16

Total probability=26

Eduardo has to pull two names out of the hat without replacing them.

Probability= Favourable outcome/Total outcome

=16/26

=8/13

For the second name:

Without replacement of the first hat

Probability of girls=16-1=15

Total probability=26-1=25

Probability= Favourable outcome/Total outcome

=15/25

=3/5

Probability of both of Eduardo's partner for the group project will be girls=8/13*3/5

=24/65

StartFraction 24 Over 65 EndFraction

4 0
3 years ago
In the game of​ roulette, a player can place a ​$9 bet on the number 11, and have a 1/38 probability of winning. If the metal ba
damaskus [11]

Answer:

1. What is the expected value of the game to the​ player?

Expected value= -$0.47368

2. If you played the game 1000​ times, how much would you expect to​ lose?

Expected value= -$473.68

Step-by-step explanation:

1. What is the expected value of the game to the​ player?

Expected value can tell how much you will gain/lose every time you do the game. To find the expected value you need to multiply every value with its chance to occur. There are two types of events here, winning and losing. You have 1/38 chance of winning $315 and 37/38 chance of losing $9.  

The expected value will be:

expected value= 1/38 * $315 + 37/38*(-$9) = $8.2894 - $8.7631

expected value= -$0.47368

You are expected to lose $0.47368 every time you play

2. If you played the game 1000​ times, how much would you expect to​ lose?

Expected value calculated for each roll. If you want to know how much you gain/lose by multiple rolls, you just need to multiply the expected value with the number of rolls. The game has negative expected value so you are expected to lose money instead of gaining money. The money you expected to lose will be:

number of roll * expected value

1000 timer * -$0.47368/time= -$473.68

You are expected to lose $473.68 if you play the game 1000 times.

6 0
3 years ago
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