Answer:
(3, 50) and (14,303)
Step-by-step explanation:
Given the system of equations;
y=23x–19 ....1
x²–y= – 6x–23 ...2
Substitute 1 into 2;
x²–(23x-19)= – 6x–23
x²–23x+19= – 6x–23 .
x²-23x + 6x + 19 + 23 = 0
x² - 17x + 42 = 0
Factorize;
x² - 14x - 3x + 42 = 0
x(x-14)-3(x-14) = 0
(x-3)(x-14) = 0
x = 3 and 14
If x = 3
y = 23(3) - 19
y = 69-19
y = 50
If x = 14
y = 23(14) - 19
y = 322-19
y = 303
Hence the coordinate solutions are (3, 50) and (14,303)
We have to calculate the probability of picking a 4 and then a 5 without replacement.
We can express this as the product of the probabilities of two events:
• The probability of picking a 4
,
• The probability of picking a 5, given that a 4 has been retired from the deck.
We have one card in the deck out of fouor cards that is a "4".
Then, the probability of picking a "4" will be:

The probability of picking a "5" will be now equal to one card (the number of 5's in the deck) divided by the number of remaining cards (3 cards):

We then calculate the probabilities of this two events happening in sequence as:

Answer: 1/12
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