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Julli [10]
2 years ago
8

Sarah had a balance of $162 in her bank account at the start of the week. She withdrew $63.50 on Monday, $23.25 on Wednesday, an

d $26.45 on Thursday. On Friday, she deposited $87.30.At the end of the week, Sarah had ___(less or more) money in her savings account than she had at the beginning of the week.
How much more or less? $_______
Mathematics
1 answer:
lozanna [386]2 years ago
8 0

Answer:

Less; $25.90 less than at the beginning of the week

Step-by-step explanation:

162-(63.50+23.25+26.45)=162-113.20=48.80

48.80+87.30=136.10; 162>136.10

162-136.10=Δ$25.90

$25.90 less

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I don’t quite understand this problem could someone help me please
ale4655 [162]

Answer:

\frac{7\sqrt{65}}{65}

Step-by-step explanation:

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3 0
2 years ago
Tessa bought stock in a restaurant for $173.00. Her stock is now worth $242.20. What is the percentage increase of the value of
frozen [14]

Answer:

The percentage increase in Tessa's stock is 40%

Step-by-step explanation:

Given that:

Previous worth = $173

Worth now = $242.20

We have to find percentage increase which can be known by subtracting previous worth from present worth and multiplying it by 100:

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i hope it will help you!

7 0
4 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
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