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givi [52]
3 years ago
12

A large grandfather clock strikes its bell once at 1:00, twice at 2:00, three times at 3:00, etcetera. what is the total number

of times the bell will be struck in a day? use an arithmetic series to help solve the problem and show how you arrived at your answer.
Mathematics
2 answers:
erastovalidia [21]3 years ago
8 0
There is a formula for adding consecutive integers.
sum = (n * (n+1)) / 2
for summing 1 through 12 it is
sum = (12 * 13) / 2 = 156/2 = 78
The clock goes through two "12-hour cycles" in a day, so the answer is 2 * 78 or 156 times.


uranmaximum [27]3 years ago
7 0

Answer:

<h2>300 times per day.</h2>

Step-by-step explanation:

Basically, to solve this problem we just have to sum:

1+2+3+...+24

But, using an arithmetic series, we have to use this formula:

Sum=n(\frac{a_{1}+a_{n} }{2})

Where, n is the total number of elements (in this case is 24), a_{1} is the first element (1) and a_{n} is the last element (24), because a day has 24 hours.

So, replacing all variables, we have:

Sum=24(\frac{1+24}{2})\\Sum=24\frac{25}{2}=300

Therefore, the bell will be struck 300 times per day.

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Answer:

19. reflection (-1, -3)

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2 years ago
The 2008 Workplace Productivity Survey, commissioned by LexisNexis and prepared by WorldOne Research, included the question, "Ho
vitfil [10]

Answer:

Therefore, the sampling distribution of \bar{x} is normal with a mean equal to 9 hours and a standard deviation of 0.7969 hours.

The 95% interval estimate of the population mean \mu is

LCL = 7.431 hours to UCL = 10.569 hours

Step-by-step explanation:

Let X be the number of hours a legal professional works on a typical workday. Imagine that X is normally distributed with a known standard deviation of 12.6.

The population standard deviation is  

\sigma = 12.6 \: hours

A sample of 250 legal professionals was surveyed, and the sample's mean response was 9 hours.

The sample size is

n = 250

The sample mean is  

\bar{x} = 9 \: hours  

Since the sample size is quite large then according to the central limit theorem, the sample mean is approximately normally distributed.

The population mean would be the same as the sample mean that is

 \mu = \bar{x} = 9 \: hours

The sample standard deviation would be  

$ s = {\frac{\sigma}{\sqrt{n} }  $

Where   is the population standard deviation and n is the sample size.

$ s = {\frac{12.6}{\sqrt{250} }  $

s = 0.7969 \: hours

Therefore, the sampling distribution of \bar{x} is normal with a mean equal to 9 hours and a standard deviation of 0.7969 hours.

The population mean confidence interval is given by

\text {confidence interval} = \mu \pm MoE\\\\

Where the margin of error is given by

$ MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) $ \\\\

Where n is the sampling size, s is the sample standard deviation and  is the t-score corresponding to a 95% confidence level.

The t-score corresponding to a 95% confidence level is

Significance level = α = 1 - 0.95 = 0.05/2 = 0.025

Degree of freedom = n - 1 = 250 - 1 = 249

From the t-table at α = 0.025 and DoF = 249

t-score = 1.9695

MoE = t_{\alpha/2}(\frac{\sigma}{\sqrt{n} } ) \\\\MoE = 1.9695\cdot \frac{12.6}{\sqrt{250} } \\\\MoE = 1.9695\cdot 0.7969\\\\MoE = 1.569\\\\

So the required 95% confidence interval is

\text {confidence interval} = \mu \pm MoE\\\\\text {confidence interval} = 9 \pm 1.569\\\\\text {LCI } = 9 - 1.569 = 7.431\\\\\text {UCI } = 9 + 1.569 = 10.569

The 95% interval estimate of the population mean \mu is

LCL = 7.431 hours to UCL = 10.569 hours

8 0
3 years ago
Which would be an equivalent way to write y= 2 - 3x ?
Fudgin [204]
The answer is B, because the numbers in the equation are 2, and -3x. The signs have to stay with the numbers they are with. So if you move the 3 to the front you have to keep the negative with it.
6 0
2 years ago
Given a dilation around the origin, what is the scale factor K? D o, K = (2,4) → (3,6)
amm1812
The scale factor is 2/3
4 0
2 years ago
Read 2 more answers
ASAP!!! Please answer!!
Ronch [10]

Answer:

x=8g/3(k+d)

Step-by-step explanation:

Move all terms to the left side and set equal to zero. Then set each factor equal to zero.

Hope this helps

6 0
2 years ago
Read 2 more answers
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