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Klio2033 [76]
2 years ago
13

Bags of sugar from a production line have a mean weight of 5.020 kg with a standard deviation of 0.078 kg. The bags of sugar are

packed in cartons of 20 bags each, and the cartons are piled in lots of 12 onto pallets for shipping. a) What percentage of cartons would be expected to contain less than 100 kg of sugar? b) Find the upper quartile of sugar content of a carton. c) What mean weight of an individual bag of sugar will result in 95% of the pallets weighing more than 1200 kg.?

Mathematics
1 answer:
Sindrei [870]2 years ago
4 0

Answer:

a.0.126

b.100.6356

c.5.010

Step-by-step explanation:

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Answer:

48

Step-by-step explanation:

8 divided by 2 = 4

4 times 12 = 48

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andrew-mc [135]
A straight line parallel to another is one that has the same slope. The generic equation of a line can be written as y-yo = m (x-xo). If we look for a parallel line that passes through the point (8.0). We should look for a line with the values m = -0.75, xo = 8, yo = 0. Substituting in the generic equation, we have that the line sought is y = -0.75 (x-8)
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2 years ago
Approximately 80,000 marriages took place in the state of New York last year. Estimate the probability that for at least one of
sasho [114]

Answer:

a)0,45119

b)1

Step-by-step explanation:

For part A of the problem we must first find the probability that both people in the couple have the same birthday (April 30)

P=\frac{1}{365} *\frac{1}{365}=\frac{1}{133225} \\

Now the poisson approximation is used

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Now, let X be the number of couples that birth April 30

P(X ≥ 1) =

1 − P(X = 0) =

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P(X ≥ 1) = 0,45119

B)  Now want to find the

probability that both partners celebrated their birthday on th, assuming that the year is 52 weeks and therefore 52 thursday

P=52*\frac{1}{365} *\frac{1}{365}=\frac{52}{133225} \\

Now the poisson approximation is used

λ=nP=80000*52/133225=31.225

Now, let X be the number of couples that birth same day

P(X ≥ 1) =

1 − P(X = 0) =

1-\frac{(e^-31.225)*(-31.225)^{0} }{0!}

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6 0
2 years ago
Solve the equation 5x + (−2) = 6x + 4 using the algebra tiles. What tiles need to be added to both sides to remove the smaller x
miss Akunina [59]

Answer:

a) Adding -5x on both sides of the equation to remove the smaller x-coefficient

b) Adding -4 on both sides will remove the constant from the right side of the equation

Step-by-step explanation:

Given equation:

5x + (−2) = 6x + 4

a) What tiles need to be added to both sides to remove the smaller x-coefficient?

Smaller x-coefficient is 5x to remove the smaller x-coefficient

So, Adding -5x on both sides of the equation to remove the smaller x-coefficient

b) What tiles need to be added to both sides to remove the constant from the right side of the equation?

the constant on right side is 4

Adding -4 on both sides will remove the constant from the right side of the equation

6 0
2 years ago
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grigory [225]
<span>247 and 248 I.e. 247 in two thetheatres and 248 in the other 4. Hence 1486</span>
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