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Irina18 [472]
3 years ago
14

Rectangle ABCD was dilated to create rectangle A'B'C'D. Triangle A B C D is dilated to form triangle A prime B prime C prime D.

Side B C is 3.8 units. Side A prime B prime is 15 units and side B prime C prime is 9.5 units. What is AB?
Mathematics
2 answers:
MArishka [77]3 years ago
8 0

Answer:

AB=6 Units

Step-by-step explanation:

Given the following lengths of rectangle ABCD and its dilation A'B'C'D' respectively

  • A'B' = 15 Units
  • B'C' = 9.5 Units
  • BC = 3.8 units

We are to determine the length of AB. To do this, we first determine the dilation factor.

D$ilation Factor =\dfrac{B'C'}{BC} =\dfrac{9.5}{3.8} =2.5

Therefore:

\dfrac{A'B'}{AB} =2.5\\\\\dfrac{15}{AB} =2.5\\$Cross multiply \\ 2.5 AB =15 \\AB=15 \div 2.5\\$Therefore, AB=6 Units

prisoha [69]3 years ago
7 0

Answer:

5 units

Step-by-step explanation:

took the test lol

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Because of staffing decisions, managers of the Gibson-Marion Hotel are interested in the variability in the number of rooms occu
olchik [2.2K]

Answer:

a) s^2 =30^2 =900

b) \frac{(19)(30)^2}{30.144} \leq \sigma^2 \leq \frac{(19)(30)^2}{10.117}

567.28 \leq \sigma^2 \leq 1690.224

c) 23.818 \leq \sigma \leq 41.112

Step-by-step explanation:

Assuming the following question: Because of staffing decisions, managers of the Gibson-Marimont Hotel are interested in  the variability in the number of rooms occupied per day during a particular season of the  year. A sample of 20 days of operation shows a sample mean of 290 rooms occupied per  day and a sample standard deviation of 30 rooms

Part a

For this case the best point of estimate for the population variance would be:

s^2 =30^2 =900

Part b

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

The degrees of freedom are given by:

df=n-1=20-1=19

Since the Confidence is 0.90 or 90%, the significance \alpha=0.1 and \alpha/2 =0.05, the critical values for this case are:

\chi^2_{\alpha/2}=30.144

\chi^2_{1- \alpha/2}=10.117

And replacing into the formula for the interval we got:

\frac{(19)(30)^2}{30.144} \leq \sigma^2 \leq \frac{(19)(30)^2}{10.117}

567.28 \leq \sigma^2 \leq 1690.224

Part c

Now we just take square root on both sides of the interval and we got:

23.818 \leq \sigma \leq 41.112

5 0
3 years ago
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