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lapo4ka [179]
3 years ago
7

Billy estimates that they will sell approximately 250 burgers thisweekend how much burger meat and fries (in pounds) should he o

rder to be prepared for this weekend
Mathematics
1 answer:
arsen [322]3 years ago
6 0

Answer:

300 Burgers

Step-by-step explanation:

Since the question gives us an approximation of 250 burgers, we can round the number to 300. This is only because they had a estimation of 250 and would like to be prepared for the weekend. Being prepared would consist of ordering more than expected to make sure the burgers do not run out.

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Write a linear function f with the values f(2) = -6 and f(0) = 2.​
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3 years ago
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The answer is 16,300. To get the answer, you just move the decimal over 4 places to the right and add your zeros in. 

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Twenty percent of drivers driving between 10 pm and 3 am are drunken drivers. In a random sample of 12 drivers driving between 1
Lesechka [4]

Answer:

(a) 0.28347

(b) 0.36909

(c) 0.0039

(d) 0.9806

Step-by-step explanation:

Given information:

n=12

p = 20% = 0.2

q = 1-p = 1-0.2 = 0.8

Binomial formula:

P(x=r)=^nC_rp^rq^{n-r}

(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.

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

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.

(d) At most 5 will be drunken drivers.

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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The answer is a(x)=x+3
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