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Romashka [77]
2 years ago
6

Determine whether the sequence converges or diverges. If it converges, find the limit. (If an answer does not exist, enter DNE.)

ln(5n) ln(15n) lim n→∞ ln(5n) ln(15n)
Mathematics
1 answer:
kenny6666 [7]2 years ago
8 0

If the limand is \ln(5n)\ln(15n), then the limit diverges. But if you mean \dfrac{\ln(5n)}{\ln(15n)}, we can do some manipulating to rewrite it as

\displaystyle\lim_{n\to\infty}\frac{\ln(5n)}{\ln(15n)}=\lim_{n\to\infty}\frac{\ln5+\ln n}{\ln15+\ln n}=\lim_{n\to\infty}\frac{\frac{\ln5}{\ln n}+1}{\frac{\ln15}{\ln n}+1}

\ln n\to\infty as n\to\infty, so the fractional terms vanish and you're left with 1.

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The XO Group Inc. conducted a survey of 13,000 brides and grooms married in the United States and found that the average cost of
slavikrds [6]

Answer:

a) 0.0392

b) 0.4688

c) At least $39,070 to be among the 5% most expensive.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 29858, \sigma = 5600

a. What is the probability that a wedding costs less than $20,000 (to 4 decimals)?

This is the pvalue of Z when X = 20000. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{20000 - 29858}{5600}

Z = -1.76

Z = -1.76 has a pvalue of 0.0392.

So this probability is 0.0392.

b. What is the probability that a wedding costs between $20,000 and $30,000 (to 4 decimals)?

This is the pvalue of Z when X = 30000 subtracted by the pvalue of Z when X = 20000.

X = 30000

Z = \frac{X - \mu}{\sigma}

Z = \frac{30000 - 29858}{5600}

Z = 0.02

Z = 0.02 has a pvalue of 0.5080.

X = 20000

Z = \frac{X - \mu}{\sigma}

Z = \frac{20000 - 29858}{5600}

Z = -1.76

Z = -1.76 has a pvalue of 0.0392.

So this probability is 0.5080 - 0.0392 = 0.4688

c. For a wedding to be among the 5% most expensive, how much would it have to cost (to the nearest whole number)?

This is the value of X when Z has a pvalue of 0.95. So this is X when Z = 1.645.

Z = \frac{X - \mu}{\sigma}

1.645 = \frac{X - 29858}{5600}

X - 29858 = 5600*1.645

X = 39070

The wedding would have to cost at least $39,070 to be among the 5% most expensive.

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2 years ago
On the way to her babysitting job, Mary rode her bike for the first mile. She stopped at a traffic light and waited a few minute
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The answer is b, hope this helps
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2 years ago
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Aluminium coated dewars are cylindrical flasks used to safely store and transport liquid nitrogen, dry ice etc. A company manufa
Nitella [24]

This question is Incomplete because it lacks the diagram showing the containers as well as the required options. Find attached to this answer the appropriate diagram.

Complete Question

Aluminium coated dewars are cylindrical flasks used to safely store and transport liquid nitrogen, dry ice etc. A company manufactures such dewars of different sizes. Shown here are two dewars, both of the same height but of different diameters. How much aluminium will be required to cover Dewar A compared to that required for Dewar B? (Note: The base and the lid are NOT made up of aluminium.)

A) 1/3 times

B) 3 times

C) 9 times

D) The same amount of aluminum will be required.

Answer:

B) 3 times

Step-by-step explanation:

From the question, we are told that we are to find the amount of aluminum to cover Dewar A and B , we are also told that the base and the lid are not covered with aluminum.

From this information, we can determine that we are to find the curved or Lateral surface area of the cylinder because the base and lid are not included

The curved surface area of a cylinder is calculated as 2πrh

We told that both cylinders have the same height. Hence,

For Dewar A

We have Diameter of 30 cm, radius = Diameter ÷ 2 = 30cm ÷ 2 = 15cm.

Height = h

The curved surface area = 2πrh

= 2 × π × 15 ×h

= 30πhcm²

For Dewar B

We have Diameter of 10 cm, radius = Diameter ÷ 2 = 10cm ÷ 2 = 5cm.

Height = h

The curved surface area = 2πrh

= 2 × π × 5 ×h

= 10πhcm²

When we compare the curved surface Dewar A to the curved surface area of Dewar B

Dewar A : Dewar B

30πhcm² : 10πhcm²

3(10πhcm²) : 10πhcm²

From the above comparison, we can see that 3 times more aluminum is required to cover Dewar A compared to Dewar B.

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In a camp there is food for 400 persons for 23 days-if 60 more persons join the camp find the number of days the provision will
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The answer is 20 days.

After 60 people have joined there will be 460 people in the camp.

The number of days which the provisions will last will be proportional less after the 60 people have joined and will be:-

(400/460) * 23

= (20 / 23) * 23

=  20


3 0
2 years ago
Round 2.1349 to the nearest hundredth. Explain in your own words why you round the number the way you do. Use a number line to s
zhenek [66]

Answer:

2.13

Step-by-step explanation:

because 2.1349 is closer to 2.13 than 2.14 because the number 4 is less than 5 so it's closer to 0

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2 years ago
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