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4 years ago

Name a rectangle that is also a square

Mathematics

2 answers:

Yuki888 [10]4 years ago

8 0

It's square a rectangle has 4 right angles just like a square. That's why the answer is square

alexdok [17]4 years ago

3 0

It can also be defined as an equiangular quadrilateral, since equiangular means that all of its angles are equal (360°/4 = 90°). It can also be defined as a parallelogram containing a right angle. A rectangle with four sides of equal length is a square. So the answer is that they are the same but the answer is SQUARE.

Hope that helps!

You might be interested in

Which number is equivalent to (1/3)^-4 A.12 B.81 C.1/12 D.1/81

marishachu [46]

Answer:

Step-by-step explanation:

(1/3)^-4

=1/[(1/3)^4]

=1/(1/81)

=81

6 0

3 years ago

In a random sample of 75 American women age 18 to 30, 26 agreed with the statement that a woman should have the right to a legal

ddd [48]

Answer:

a) $z=\frac{0.347-0.328}{\sqrt{0.338(1-0.338)(\frac{1}{75}+\frac{1}{64})}}=0.236$

$p_v =2*P(Z>0.236)=0.813$

If we compare the p value and using any significance level for example $\alpha=0.01$ always $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say that we don't have significant differences between the two proportions.

b) We are confident at 99% that the difference between the two proportions is between $-0.188 \leq p_B -p_A \leq 0.226$

Step-by-step explanation:

Previous concepts and data given

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".

$p_A$ represent the real population proportion of women age 18 to 30 agreed with the statement that a woman should have the right to a legal abortion for any reason

$\hat p_A =\frac{26}{75}=0.347$ represent the estimated proportion of women age 18 to 30 agreed with the statement that a woman should have the right to a legal abortion for any reason

$n_A=75$ is the sample size for A

$p_B$ represent the real population proportion for women age 58 to 70 agreed with the statement that a woman should have the right to a legal abortion for any reason

$\hat p_B =\frac{21}{64}=0.328$ represent the estimated proportion of women age 58 to 70 agreed with the statement that a woman should have the right to a legal abortion for any reason

$n_B=64$ is the sample size required for B

$z$ represent the critical value for the margin of error and for the statisitc

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Part a

We need to conduct a hypothesis in order to check if the proportion are equal, the system of hypothesis would be:

Null hypothesis: $p_{A} = p_{B}$

Alternative hypothesis: $p_{A} \neq p_{B}$

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{A}-p_{B}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{A}}+\frac{1}{n_{B}})}}$ (1)

Where $\hat p=\frac{X_{A}+X_{B}}{n_{A}+n_{B}}=\frac{26+21}{75+64}=0.338$

Calculate the statistic

Replacing in formula (1) the values obtained we got this:

$z=\frac{0.347-0.328}{\sqrt{0.338(1-0.338)(\frac{1}{75}+\frac{1}{64})}}=0.236$

Statistical decision

Since is a two sided test the p value would be:

$p_v =2*P(Z>0.236)=0.813$

Part b

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$(0.347-0.328) - 2.58 \sqrt{\frac{0.347(1-0.347)}{75} +\frac{0.328(1-0.328)}{64}}=-0.188$

$(0.347-0.328) + 2.58 \sqrt{\frac{0.347(1-0.347)}{75} +\frac{0.328(1-0.328)}{64}}=0.226$

And the 99% confidence interval for the difference of proportions would be given (-0.188;0.226).

We are confident at 99% that the difference between the two proportions is between $-0.188 \leq p_B -p_A \leq 0.226$

5 0

3 years ago

What’s the domain and range

Softa [21]

Answer:

domain is x range is y

Step-by-step explanation:

4 0

3 years ago

Find the (a) mean, (b) median, (c) mode, and (d) midrange for the data and then (e) answer the given question. Listed below

mafiozo [28]

Answer:

a) $\bar X = 369.62$

b) $Median=175$

c) $Mode =450$

With a frequency of 4

d) $MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

e) $s = 621.76$

And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

Step-by-step explanation:

We have the following data set given:

49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000

Part a

The mean can be calculated with this formula:

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

Replacing we got:

$\bar X = 369.62$

Part b

Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:

$Median=175$

Part c

The mode is the most repeated value in the sample and for this case is:

$Mode =450$

With a frequency of 4

Part d

The midrange for this case is defined as:

$MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

Part e

For this case we can calculate the deviation given by:

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

And replacing we got:

$s = 621.76$

And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

5 0

3 years ago

Pls help giving brainlyist

dangina [55]

Answer:

x = 0

Step-by-step explanation:

when y is -3, x is 0

5 0

3 years ago