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erma4kov [3.2K]
2 years ago
14

Look at this set of ordered pairs:

Mathematics
1 answer:
cestrela7 [59]2 years ago
3 0

Answer:

Yes!

Step-by-step explanation:

There are no repeating y values or x values!!

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I need help with this
34kurt

Answer:

it would be (-8,14)

Step-by-step explanation:

so if you plot it on a graph all ypu would do is flip it over the x axis

4 0
3 years ago
Solve for a r=x(a+b)<br><br><br> Solve for x x/2=-7
Anastaziya [24]

Answer:

a = (r/x) -b

x = -14

Step-by-step explanation:

r=x(a+b)

Divide each side by x

r/x = a+b

Subtract b

r/x  -b = a

x/2=-7

Multiply each side by 2

x/2*2 = -7*2

x = -14

4 0
2 years ago
Read 2 more answers
Consider the equation y = –3(x – 3) Which equation, when graphed with the given equation, will form a system with lines that ove
pentagon [3]

Answer:

y=-3x+9 (D)

Step-by-step explanation:

8 0
3 years ago
A dinner for two at the local seafood restaurant cost
Vlad [161]

Answer:

63.38

Step-by-step explanation:

(49.90)(20%)(6%)

20%= $9.98

49.90+9.98 =59.80

(59.80)(6%)

6%= $3.58

59.80+3.58=63.38

8 0
2 years ago
A group of 30 students from your school is part of the audience for a TV game show. The total number of people in the audience i
s344n2d4d5 [400]

Answer:

5.3%

Step-by-step explanation:

The final probability is calculated by means of the quotient of the specific combinations and the total of total combinations

Let's start with the specific ones,

First the number of combinations of 4 of the 30 students getting a spot, i.e .:

A combinations are equal to:

nCx = n! / x! * (n-x)!

Replacing:

30C4 = 30! / (4! * 26!) = 27405

Segundo the number of combinations of the other audience members filling the other 4 (8-4) spots n = 110, 140 - 30

110C4 = 110! / (4! * 106!) = 5773185

Now the total combinations of possible 8 contestants from the audience

140C8 = 140! / (8! * 132!) = 2.98 * 10 ^ 12

Finally, the probability is equal to:

P = (30C4 * 110C4) / 140C8

replacing:

P = 27405 * 5773185 / 2.98 * 10 ^ 12

P = 0.053

Therefore the probability is 5.3%

4 0
3 years ago
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