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2 years ago

Which pairs of rectangles are similar polygons? Select each correct answer.

Mathematics

2 answers:

oee [108]2 years ago

8 0

The corrects answers are B and D for this one.

Since you divide the same side for both the the rectangles on both parts and if the division statements are equaled, then they are similar.

For example;

Answer B.
50 / 10 = 5
and
30 / 6 = 5

Answer D.
7.5 / 2.5 = 3
and
4.5 / 1.5 = 3

Since they both equal the same for both of the rectangles for the answers individually, they are similar polygon rectangles.

Hope this helps!

~ ShadowXReaper069

Andrews [41]2 years ago

3 0

Answer: Second and fourth pairs of rectangle are similar.

Explanation: Since, If two Shapes are similar then the ratios of their corresponding sides are equal.

In First figure ratios of corresponding sides are not equal.

Because, $\frac{100}{10}$ ≠ $\frac{10}{7}$

Therefore, they are not similar.

In second figure ratios of corresponding sides are equal.

Because, $\frac{50}{10}$ = $\frac{30}{6}=5$

Therefore, they are similar.

In third figure ratios of corresponding sides are not equal.

Because, $\frac{5.2}{2.6}$ ≠ $\frac{3.8}{1.2}$

Therefore, they are not similar.

In fourth figure ratios of corresponding sides are equal.

Because, $\frac{7.5}{2.5}$ = $\frac{4.5}{1.5}=3$

Therefore, they are similar.

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a) $\bar X = \frac{\sum_{i=1}^n X_i}{n}$

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The variance for this estimator is given by:

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We can assume the obervations independent so then we have:

$Var(\frac{x_{8} +x_{9}}{2}) = \frac{1}{4} (2\sigma^2) = \frac{\sigma^2}{2}$

And replacing we got:

$Var(\frac{x_{8} +x_{9}}{2})= \frac{0.3105^2}{2}= 0.0482$

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Step-by-step explanation:

Data given:

0.86 0.88 0.88 1.07 1.09 1.17 1.29 1.31 1.46 1.49 1.59 1.62 1.65 1.71 1.76 1.83

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And for this case if we use this formula we got:

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Part b

For this case in order to calculate the median we need to put the data on increasing way like this:

0.86 0.88 0.88 1.07 1.09 1.17 1.29 1.31 1.46 1.49 1.59 1.62 1.65 1.71 1.76 1.83

Since we have n =16 values for the sample the median can be calculated as the average between position 8th anf 9th and we got:

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Part c

For this case we can assume that the mean is $\mu = 1.3538$

And we can calculate the population deviation with the following formula:

$\sigma = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{N}}$

And if we replace we got: $\sigma= 0.3105$

And assuming normal distribution we have this:

$X \sim N (\mu = 1.3538, \sigma= 0.3105)$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.1$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.28$

And if we solve for a we got

$a=1.3538 +1.28*0.3505=1.8024$

So the value of height that separates the bottom 90% of data from the top 10% is 1.8024.

Part d

The median is defined as :

$Median= \frac{x_{8} +x_{9}}{2}$

The variance for this estimator is given by:

$Var(\frac{x_{8} +x_{9}}{2}) = \frac{1}{4} Var(X_{8} +X_{9})$

We can assume the obervations independent so then we have:

$Var(\frac{x_{8} +x_{9}}{2}) = \frac{1}{4} (2\sigma^2) = \frac{\sigma^2}{2}$

And replacing we got:

$Var(\frac{x_{8} +x_{9}}{2})= \frac{0.3105^2}{2}= 0.0482$

And the standard error would be given by:

$Sd(\frac{x_{8} +x_{9}}{2})= \sqrt{0.0482}=0.2196$

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2 years ago