Given:
Budget = $30
cost of pizza = $9
cost of drinks = $1
number of pizza = p
number of drinks = d
pizza and drinks should only be equal to or less than the budget of $30.
$9p + $1d <u>< </u>$30
unit cost of pizza : 9
unit cost of drinks : 1
total cost of set (p & d) 10
pizza = 9/10 x 30 = 27 total cost of pizza
drinks =1/10 x 30 = 3 total cost of drinks
27 ÷ 9 = 3 number of pizzas to order
3 ÷ 1 = 3 number of drinks to order
To check:
9p + 1d <u>< </u>30
9(3) + 1(3) <u>< </u>30
27 + 3 <u>< </u>30
30 <u>< </u>30
Answer:
No one cares about work
Step-by-step explanation:
Answer:
Solve by adding or subtracting like terms. (Ex. 2k + 4k = 6k; 9 + 7 = 16)
- 6k + 7k = 1k = k
(7 - 6 = 1)
12r - 8 - 12 = 12r - 20
(8 + 12 = 20 but sign is negative) 12r will stay the same because there are no terms with a letter r.
n - 10 + 9n - 3 = 9n + n - 10 - 3 = 10n - 10 - 3 = 10n - 13
(n = 1 so 9 + 1 = 10; 10 + 3 = 13 but negative sign)
- 4x - 10x = 14x
(10 + 4 = 14)
- r - 10r = 11r
(r = 1 so 10 + 1 = 11)
Intuitively, one would think the ball would land in the green spot 2 out of the 38 times, since there are 38 slots and 2 are green.
The probability that it lands in a green section is 2/38. Multiplying this by the number of times the experiment is performed, we get (2/38)(38) = 2.
