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tatuchka [14]
3 years ago
10

A snack bar sells two sizes of snack packs. A large snack pack costs 5 dollars. A small snack pack costs 3 dollars. In one day t

he snack bar sells 60 snack packs for a total of 220 dollars how many small snacks packs did they sell.
Mathematics
1 answer:
il63 [147K]3 years ago
7 0
X=cost of large snack Y=cost of small snack X+Y=60 5X+3Y=220 To solve for y, solve x+y=60 for x. Subtract y from both sides so you're left with x=-y+60. You now plug in -y+60 for x in the second equation. 5(-y+60)+3y=220 -5y+300+3y=220 -2y+300=220 -2y=-80 Y=40 40 small snack packs is the final answer
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q = 1-p = 1-0.2 = 0.8

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(a) Exactly two will be drunken drivers.

P(x=2)=^{12}C_{2}(0.2)^{2}(0.8)^{12-2}

P(x=2)=66(0.2)^{2}(0.8)^{10}

P(x=2)=\approx 0.28347

Therefore, the probability that exactly two will be drunken drivers is 0.28347.

(b)Three or four will be drunken drivers.

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P(x=3\text{ or }x=4)=P(x=3)+P(x=4)

Using binomial we get

P(x=3\text{ or }x=4)=^{12}C_{3}(0.2)^{3}(0.8)^{12-3}+^{12}C_{4}(0.2)^{4}(0.8)^{12-4}

P(x=3\text{ or }x=4)=0.236223+0.132876

P(x=3\text{ or }x=4)\approx 0.369099

Therefore, the probability that three or four will be drunken drivers is 0.3691.

(c)

At least 7 will be drunken drivers.

P(x\geq 7)=1-P(x

P(x\leq 7)=1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)+P(x=6)]

P(x\leq 7)=1-[0.06872+0.20616+0.28347+0.23622+0.13288+0.05315+0.0155]

P(x\leq 7)=1-[0.9961]

P(x\leq 7)=0.0039

Therefore, the probability of at least 7 will be drunken drivers is 0.0039.

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P(x\leq 5)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)

P(x\leq 5)=0.06872+0.20616+0.28347+0.23622+0.13288+0.05315

P(x\leq 5)=0.9806

Therefore, the probability of at most 5 will be drunken drivers is 0.9806.

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3 years ago
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