Answer:
- 197 popcorn bags
- 202 food trays
Step-by-step explanation:
Since the numbers are split evenly 5 ways, each stand gets 1/5 of the total:
(985 bags)/5 = 197 bags
(1010 trays)/5 = 202 trays
Each concession stand receives 197 popcorn bags and 202 food trays.
Let's
simplify step-by-step.
3x2(x2−4x−4)+5x3+7x2+2x+11
Distribute:
=(3x2)(x2)+(3x2)(−4x)+(3x2)(−4)+5x3+7x2+2x+11
=3x4+−12x3+−12x2+5x3+7x2+2x+11
Combine
Like Terms:
=3x4+−12x3+−12x2+5x3+7x2+2x+11
=(3x4)+(−12x3+5x3)+(−12x2+7x2)+(2x)+(11)
=3x4+−7x3+−5x2+2x+11
Answer:
=3x4−7x3−5x2+2x+11
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Step-by-step explanation:
(1 + cos θ + sin θ) / (1 + cos θ − sin θ)
Multiply by the reciprocal:
(1 + cos θ + sin θ) / (1 + cos θ − sin θ) × (1 + cos θ + sin θ) / (1 + cos θ + sin θ)
(1 + cos θ + sin θ)² / [ (1 + cos θ − sin θ) (1 + cos θ + sin θ) ]
(1 + cos θ + sin θ)² / [ (1 + cos θ)² − sin² θ) ]
Distribute and simplify:
(1 + cos θ + sin θ)² / (1 + 2 cos θ + cos² θ − sin² θ)
[ 1 + 2 (cos θ + sin θ) + (cos θ + sin θ)² ] / (1 + 2 cos θ + cos² θ − sin² θ)
(1 + 2 cos θ + 2 sin θ + cos² θ + 2 sin θ cos θ + sin² θ) / (1 + 2 cos θ + cos² θ − sin² θ)
Use Pythagorean identity:
(2 + 2 cos θ + 2 sin θ + 2 sin θ cos θ) / (sin² θ + cos² θ + 2 cos θ + cos² θ − sin² θ)
(2 + 2 cos θ + 2 sin θ + 2 sin θ cos θ) / (2 cos² θ + 2 cos θ)
(1 + cos θ + sin θ + sin θ cos θ) / (cos² θ + cos θ)
Factor:
(1 + cos θ + sin θ (1 + cos θ)) / (cos θ (1 + cos θ))
(1 + cos θ)(1 + sin θ) / (cos θ (1 + cos θ))
(1 + sin θ) / cos θ
10^10 = 100
20^2 = 400
29^29 = 841
28^28 = 784
Answer:
X ____ 0 ______ 1 _______ 2
P(x) __ 1/7 _____ 4/7 _____ 2/7
Step-by-step explanation:
Given that :
Number of red balls = 4
Number of Blue balls = 3
Total number of balls = (4 + 3) = 7
Picking without replacement :
Number of picks = 2
1st pick is red :
Probability = (required outcome / Total possible outcomes)
USing combination formula :
nCr = n! / (n-r)! r!
Total possible outcome :
7C2 = 7! / 5!2! = 7*6 / 2 = 42 / 2 = 21
X = number of red
For red = 0;
P(x =0) = required outcome / Total possible outcomes
P(x = 0)
Required outcome : [4C0 * 3C2)]
P(x = 0) = [(4C0 * 3C2)] / 21 = (1*3)/21 = 3/21 = 1/7
P(x = 1)
Required outcome : [4C1 * 3C1)]
P(x = 1) = [(4C1 * 3C1)] / 21 = (4*3)/21 = 12/21 = 4/7
P(x = 2)
Required outcome : [4C2 * 3C0)]
P(x = 2) = [(4C2 * 3C0)] / 21 = (6*1)/21 = 6/21 = 2/7
Hence ;
X ____ 0 ______ 1 _______ 2
P(x) __ 1/7 _____ 4/7 _____ 2/7
