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Dmitrij [34]
3 years ago
5

X = 4. 5x +23x^3 - 360.

Mathematics
2 answers:
belka [17]3 years ago
6 0

Answer:

1132

Step-by-step explanation:

if x=4 than all you need to do is replace 'x' with '4'

5(4)+23(4)^3-360

20+23(64)-360

20+1472-360

1492-360

1132

Zinaida [17]3 years ago
4 0

Missing: 360 ‎| ‎Must include: ‎360

look at

https://www.google.com/search?q=x+%3D+4.+5x+%2B23x%5E3+-+360&rlz=1CAJAIV_enUS838&oq=x+%3D+4.+5x+%2B23x%5E3+-+360&aqs=chrome..69i57&sourceid=chrome&ie=UTF-8&safe=active&ssui=on

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Samuel spends a winter day recording the temperature once every three hours for science class. At 9 am, the temperature was -14.
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Answer:

The temperature at 9 am = -14.7°F

Between 9am and 12 pm = -14.7 + 17.8 = 3.1°F

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The figure represents the shape of a kitchen floor. What is the area? (round to the nearest hundredth)
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We would divide the shape into a rectangle and a trapezium

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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

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