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Fed [463]
3 years ago
12

PLEASE HELPPPPPPPP ASAAAAP1!!!!!!!!

Mathematics
1 answer:
Radda [10]3 years ago
3 0
As far as I can tell, that isn't a function at all. 

x = 0 produces y = -2 and y = 4, which can't happen.

Please don't use all caps, that's really, really irritating.
You might be interested in
Wich is the answer of 7+5×6​
solniwko [45]

Answer:

37

Step-by-step explanation:

Using order of operations, multiplication is done before addition.

given

7 + 5 × 6 ← perform multiplication

= 7 + 30 ← do the addition

= 37

8 0
3 years ago
Read 2 more answers
Help me please will brainlist
Thepotemich [5.8K]

Answer:B

Step-by-step explanation:

18 is greater than >16

18> 8-8(-1)

18>8+8

8 0
3 years ago
Because gambling is a big​ business, calculating the odds of a gambler winning or losing in every game is crucial to the financi
IRISSAK [1]

Answer:

a) 0.1165

b) 0.0983

c) 0.000455

d) 0.787

e) 0.767

Step-by-step explanation:

5 ​bars, 4 ​lemons, 3 ​cherries, and a bell

Total = 5+4+3+1 = 13

The probability of getting a bar on a slot, P(Ba) = 5/13 = 0.385

A lemon, P(L) = 4/13 = 0.308

A cherry, P(C) = 3/13 = 0.231

A bell, P(Be) = 1/13 = 0.0769

a) Probability of getting 3 lemons = (4/13) × (4/13) × (4/13) = 256/2197 = 0.1165

b) Probability of getting no fruit symbol

On each slot, there are 4+3 = 7 fruit symbols.

Probability of getting a fruit symbol On a slot = 7/13

Probability of not getting a fruit symbol = 1 - (7/13) = 6/13 = 0.462

Probability of not getting a fruit symbol On the three slots = 0.462 × 0.462 × 0.462 = 0.0983

c) Probability of getting 3 bells, the jackpot = (1/13) × (1/13) × (1/13) = 1/2197 = 0.000455

d) Probability of not getting a bell on the 3 slots

Probability of not getting a bell on one slot = 1 - (1/13) = 12/13 = 0.923

Probability of not getting a bell on the 3 slots = (12/13) × (12/13) × (12/13) = 1728/2197 = 0.787

e) Probability of at least one bar is a sum of probabilities

Note that Probability of getting a bar = 5/13 and probability of not getting a bar = 8/13

1) Probability of getting 1 bar and other stuff on the 2 other slots (this can happen in 3 different orders) = 3 × (5/13)×(8/13)×(8/13) = 960/2197 = 0.437

2) Probability of getting 2 bars and other stuff on the remaining slot (this can also occur in 3 different orders) = 3 × (5/13)×(5/13)×(8/13) = 600/2197 = 0.273

3) Probability of getting 3 bars on the slots machine = (5/13) × (5/13) × (5/13) = 125/2197 = 0.0569

Probability of at least one bar = 0.437 + 0.273 + 0.0569 = 0.7669 = 0.767

5 0
3 years ago
What is -23/6 - 7/3<br>A-37/6<br>B-30/9<br>C-16/6<br>D37/6​
8_murik_8 [283]

Answer:

A

Step-by-step explanation:

i solved it

7 0
3 years ago
2sec^2x - secx - 1 = 0
Ivenika [448]
What are you trying to do here?
Solve the graph, or make it appear as something else?

First, we're going to take one sec (x) out so that we get:
sec (x) (2sec (x) -1 -1) = 0

sec (x) (2sec (x) -2) = 0


Then we're going to separate the two to find the zeros of each because anything time 0 is zero.

sec(x) = 0

2sec (x) - 2 = 0

Now, let's simplify the second one as the first one is already.

Add 2 to both sides:

2sec (x) = 2

Divide by 3 on both sides:

sec (x) = 1


I forgot my unit circle, so you'd have to do that by yourself. Hopefully, I helped a bit though!
8 0
3 years ago
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