82. Dora bought one package of each 1-pound pork, 2-pound pork and 4-pound pork.
Thus, she got a total of:
=> 1 pound + 2 pounds + 4 pounds = 7 pounds of pork.
Question: How many ¼ pound of hamburger she can make then with this given number of pork in pounds.
=> ¼ = 1 / 4 = .25
Now, let’s divide 7 pounds by .25 pounds
=> 7 / .25 = 28
Thus, She can make 28 hamburgers in all.
Answer:
A
Step-by-step explanation:
-75=-25a
Divide both sides by -3:
3=a
a=3
The answer is choice A
27/20
= 20/20 + 7/20
= 1 + 7/20
= 1.00 + 0.35
= 1.35
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List of values to remember (extras):
0.05 = 1/20 = 5%
0.10 = 1/10 = 10%
0.15 = 3/20 = 15%
0.20 = 1/5 = 20%
0.25 = 1/4 = 25%
0.30 = 3/10 = 30%
0.35 = 7/20 = 35%
0.40 = 2/5 = 40%
0.45 = 9/20 = 45%
0.50 = 1/2 = 50%
0.55 = 11/20 = 55%
0.60 = 3/5 = 60%
0.65 = 13/20 = 65%
0.70 = 7/10 = 70%
0.75 = 3/4 = 75%
0.80 = 4/5 = 80%
0.85 = 17/20 = 85%
0.90 = 9/10 = 90%
0.95 = 19/20 = 95%
1.00 = 1 = 100%
When you have 3 choices for each of 6 spins, the number of possible "words" is
3^6 = 729
The number of permutations of 6 things that are 3 groups of 2 is
6!/(2!×2!×2!) = 720/8 = 90
A) The probability of a word containing two of each of the letters is 90/729 = 10/81
The number of permutations of 6 things from two groups of different sizes is
(2 and 4) : 6!/(2!×4!) = 15
(3 and 3) : 6!/(3!×3!) = 20
(4 and 2) : 15
(5 and 1) : 6
(6 and 0) : 1
B) The number of ways there can be at least 2 "a"s and no "b"s is
15 + 20 + 15 + 6 + 1 = 57
The probability of a word containing at least 2 "a"s and no "b"s is 57/729 = 19/243.
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These numbers were verified by listing all possibilities and actually counting the ones that met your requirements.
Question 1 demonstrates the Commutative Property.