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Veseljchak [2.6K]

3 years ago

15

An irregular parallelogram rotates 360° about the midpoint of its diagonal. How many times does the image of the parallelogram c

oincide with its preimage during the rotation?

1 answer:

Helga [31]3 years ago

6 0

2 times it will coincide

You might be interested in

Donnell and Maria are both members of a populationand a simple random sample is being conducted. If the chance of Donnell being

harina [27]

Answer:

289/290

Step-by-step explanation:

Given that the chance of Donnell being selected is 1/290

then P( Donnell being selected )= 1/290

Since Donnell and Maria are both member of a population.

from probability theorem which gives the likely hood of an event to happen which cannot be more than 1, then the probability of Donnell and Maria to be selected is 1

P(Donnell and Maria)= 1

But the chance of Donnell being selected is 1/290

Then,

1/290 + P(Maria)= 1

P(Maria)= 1-(1/290)

= 289/290

the chance of Maria being selected is 289/290

4 0

3 years ago

Write (n+4)(n+3)(n+2) as a permutation

kenny6666 [7]

Answer:

<h2>1267 456 578 123 679 333 5</h2>

Step-by-step explanation:

yo

7 0

3 years ago

Below are survival times (in days) of 13 guinea pigs that were injected with a bacterial infection in a medical study:

tresset_1 [31]

Outliers are data that are relatively far from other data elements.

The dataset has an outlier and the outlier is 120

The dataset is given as:

91 83 84 79 91 93 95 97 97 120 101 105 98

Sort the dataset in ascending order

79 83 84 91 91 93 95 97 97 98 101 105 120

<h3>The lower quartile (Q1)</h3>

The Q1 is then calculated as:

$Q1 = \frac{N +1}{4}th$

So, we have:

$Q1 = \frac{13 +1}{4}th$

$Q1 = \frac{14}{4}th$

$Q1 = 3.5th$

This is the average of the 3rd and the 4th element

$Q1 = \frac{1}{2} \times (84 + 91)$

$Q1 = 87.5$

<h3>The upper quartile (Q3)</h3>

The Q3 is then calculated as:

$Q3 = 3 \times \frac{N +1}{4}th$

So, we have:

$Q3 = 3 \times \frac{13 +1}{4}th$

$Q3 = 3 \times 3.5th$

$Q3 = 10.5th$

This is the average of the 10th and the 11th element.

$Q_3 =\frac12 \times (98 + 101)$

$Q_3 =99.5$

<h3>The interquartile range (IQR)</h3>

The IQR is then calculated as:

$IQR = Q_3 -Q_1$

$IQR = 99.5 - 87.5$

$IQR = 12$

Also, we have:

$IQR(1.5) = 12 \times 1.5$

$IQR(1.5) = 18$

<h3>The outlier range</h3>

The lower and the upper outlier range are calculated as follows:

$Lower = Q_1 - IQR(1.5)$

$Lower = 87.5- 18$

$Lower = 69.5$

$Upper = Q_3 + IQR(1.5)$

$Upper = 99.5 + 18$

$Upper = 117.5$

120 is greater than 117.5.

Hence, the dataset has an outlier and the outlier is 120

Read more about outliers at:

brainly.com/question/9933184

6 0

2 years ago

10 + (2^2 + 3^2) x 2<br><br><br> PEMDAS

yarga [219]

The answer would be 36.

Following PEMDAS, you'd solve what's in the parenthesis first. 2^2 = 4 and 3^2 = 9. This leaves you with 10 + (4 + 9) x 2. Add 9 and 4 to get 13, and then the parenthesis are done.

Next, you'd multiple 13 and 2 (since it's the next step in PEMDAS) to get 26.

Finally, you add together 10 and 26 to get the answer, which is 36.

I hope this helps!

3 0

3 years ago

Read 2 more answers

Please help me with this problem! Also, please stop posting scam links under my questions.

almond37 [142]

Answer:

the second option is correct, the one that you picked

Step-by-step explanation:

6 0

2 years ago