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Anvisha [2.4K]
3 years ago
14

Due to renovations at her workplace, Lucía must move from her current office into a temporary office. Her current office is a sq

uare room of dimension xx feet. Her temporary office will be a rectangular room with a length 33 feet greater than her current office's length and a width 55 feet less than her current office's width. Which of the following expressions gives the difference between the area, in square feet, of Lucía's current office and the area, in square feet, of Lucía's temporary office?
A
2x+152x+15

B
2x-152x−15

C
-2x+15−2x+15

D
-8x-15−8x−15

E
8x+158x+15
Mathematics
2 answers:
Amanda [17]3 years ago
4 0

Step-by-step explanation:

current \: office \: area = x \times x = x {}^{2}. \\ temporary \: office \: dimensions \\  = x  +  3 \: and \: x - 5. \\ current \: office \: area \\  = (x + 3)(x - 5) = x {}^{2}  - 2x + 15 \\  \\ x {}^{2} (x {}^{2}  - 2x + 15) \\  = x {}^{2}  - x {}^{2}  - 2x + 15 \\  = 2x + 15. \\ ans(2x + 15).

Oksi-84 [34.3K]3 years ago
3 0

Answer:

  • A. 2x + 15

Step-by-step explanation:

<u>Current office area:</u>

  • x*x = x²

<u>Temporary office dimensions:</u>

  • x + 3 and x - 5

<u>Temporary office area:</u>

  • (x + 3)(x - 5) = x² - 2x - 15

<u>The difference is:</u>

  • x² - (x²- 2x - 15) =
  • x² - x² + 2x + 15 =
  • 2x + 15

Correct choice is A

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The following data were collected from 12 rain gauges in a park. Build a 95% CI for the mean rainfall at the park.
dybincka [34]

Answer:

Critical values:t_{\alpha/2}=-2.201 t_{1-\alpha/2}=2.201

95% confidence interval would be given by (3.646;4.472)

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The data is:

4.65 3.89 2.73 4.35 3.80 4.86 4.33 4.37 4.76 4.05 3.05 3.87

2) Compute the sample mean and sample standard deviation.

In order to calculate the mean and the sample deviation we need to have on mind the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}

=AVERAGE(4.65,3.89,2.73, 4.35, 3.8, 4.86, 4.33, 4.37, 4.76, 4.05, 3.05, 3.87)

On this case the average is \bar X= 4.059

=STDEV.S(4.65,3.89,2.73, 4.35, 3.8, 4.86, 4.33, 4.37, 4.76, 4.05, 3.05, 3.87)

The sample standard deviation obtained was s=0.6503

3) Find the critical value t* Use the formula for a CI to find upper and lower endpoints

In order to find the critical value we need to take in count that our sample size n =12 <30 and on this case we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. The degrees of freedom are given by:

df=n-1=12-1=11

We can find the critical values in excel using the following formulas:

"=T.INV(0.025,11)" for t_{\alpha/2}=-2.201

"=T.INV(1-0.025,11)" for t_{1-\alpha/2}=2.201

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}  

And we can use Excel to calculate the limits for the interval

Lower interval : "=4.059 -2.201*(0.6503/SQRT(12))" =3.646

Upper interval :  "=4.059 +2.201*(0.6503/SQRT(12))" =4.472

So the 95% confidence interval would be given by (3.646;4.472)

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