For the first question, all you need to know is that the (amount you want) divided by (the amount of cards total) is the probability of getting the amount you want at random. for example, if there are 3 red marbles in a bag of ten marbles, you divide 3 by 10 (0.30) the probability of picking up a red marble is 30%. try adding up how many cards you want to pick up by the number of cards total.
9514 1404 393
Answer:
(3) y = (x -3)²
Step-by-step explanation:
Try the first value of x in each function:
(1) y = 3² -3 ≠ 0
(2) y = (3 +3)² ≠ 0
(3) y = (3 -3)² = 0 . . . . . . matches the table value for f(3)
(4) y = 3² +3 ≠ 0
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The only viable choice is (3).
Set up
Let the dimes = d
Let the pennies = p
Let the quarters = q
Equations
You cannot mean that the pennies and dimes have equal numbers. That would mean that each had 21.5 members. Now could you mean that the dime and penny amount could be the same with 43 coins that total 4.00. Four dollars means that you need 40 dimes alone. It must mean that you are including quarters.
p + d + p = 43 (1)
p = d (2)
p +10d +25q = 451 (3)
Note how this last equation = was derived. You have to multiply the dimes by 10 and he quarters by 100 and the total by 100 to get the numbers all in pennies.
Put the results of 2 into 1.
2p + q = 43 (4)
You need to modify equation 3 as well.
p + 10p + 25q = 451
11p + 25q = 451 (5)
Solve the new equations
2p + q = 43 (4)
11p + 25q = 451 (5)
Multiply 4 by 25
25(2p- + q = 43)
50p + 25q = 1075 (6) Subtract (5) from (6)
<u>11p + 25q = 451
</u>39p = 624 Divide by 39
p = 624 / 39
p = 16
Since the pennies and dimes are equal there are 16 dimes
p + d + q = 43
16 + 16 + q = 43
32 + q = 43
q = 11
Check
16 + 10*16 + 11*25 = ?
16 + 160 + 275 = ?
451 = ?
Nice problem. Thanks for posting.
