There were seven hundred eight employees working in an office building seventy one of them left to go have lunch how

2 answers:

6 0

Okay so let's break this down.

Step 1: First thing we want to do is highlight the major words, the highlighted words are below:

seven hundred eight
seventy one

Step 2: Let's write the equation out

So we know that there are seven hundred eight employees and that is 708
We know that seventy-one of them left and that is 71

Step 3: Make an equation
If there was a total of 708 and 71 left the equation would be 708 - 71 = how many employees are left.

Step 4: solve equation
708 - 71 = 637

There for there are 637 employees left working in and office.
Hope this helps - Alyssa :)

(P.S It would mean a lot to me if you marked my answer Brainliest answer, I am trying to get a higher ranking. Thank you!)

Anarel [89]3 years ago

3 0

708-71= 637

There will be 637 employees left.

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Suppose a sample of 100 families of four vacationing at Niagara Falls resulted in sample mean of $282.45 spent per day and a sam

Juli2301 [7.4K]

Given Information:

Mean = μ = $282.45

Standard deviation = σ = $64.50

Sample size = n = 100

Confidence level = 95%

Required Information:

95% Confidence interval = ?

Answer:

95% Confidence interval = ($269.81, $295.09)

Step-by-step explanation:

We are given a Normal Distribution, which is a continuous probability distribution and is symmetrical around the mean. The shape of this distribution is like a bell curve and most of the data is clustered around the mean. The area under this bell shaped curve represents the probability.

What is Confidence Interval?

The confidence interval represents an interval that we can guarantee that the target variable will be within this interval for a given confidence level.

The confidence interval is given by

$CI = \bar{x} \pm MoE\\$

Where $\bar{x}$ is the mean and MoE is the margin of error given by

$MoE = z_{\alpha/2}(\frac{\sigma}{\sqrt{n} } ) \\$

Where σ is the standard deviation, n is the sample size and $z_{\alpha/2}$ is the z-score corresponding to 95% confidence level.

$z_{\alpha/2} = 1 - 0.95 = 0.05/2 = 0.025\\\\z_{0.025} = 1.96$

$MoE = 1.96\cdot \frac{64.50}{\sqrt{100} } \\\\MoE = 1.96\cdot 6.45\\\\MoE = 12.64\\$

Finally, the confidence interval is

$CI = \bar{x} \pm MoE\\\\CI = 282.45 \pm 12.64\\\\CI = 282.45 - 12.64 \: and \: 282.45 + 12.64\\\\CI = \$269.81 \: and \:\:\$295.09\\$

Therefore, we are 95% sure that the true population mean amount spent per day by a family of four visiting Niagara Falls is within the interval of ($269.81, $295.09)

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