Ask question

Ask question

All categories

ArbitrLikvidat [17]

3 years ago

8

The figure shows a parallelogram inside a rectangle outline:

1 answer:

madreJ [45]3 years ago

7 0

Ok First put in you're question 1 2 3 4 the answer is 1 over 5

You might be interested in

If f(x)= 3/x+2 - square root of x-3, the domain for f(x) is all real numbers ___ than or equal to 3

aksik [14]

Answer:

I don't get the question very well

5 0

3 years ago

Read 2 more answers

In the diagram below, BE ~ BC, mLC = 32° and m/BFD = 103°. Find<br> m

steposvetlana [31]

Answer:

Here's a rainbow of logic.

Step-by-step explanation:

5 0

2 years ago

Free brainleist What is the product of five square root five times six square root four ? Simplify your answer. (5 points)

SSSSS [86.1K]

Sixty square root five (simplified) There ya go hope it helps

6 0

3 years ago

Read 2 more answers

During rush hour, Geoffrey counted that he had to stop at 8 different red lights. He went through a total of 25 intersections. W

omeli [17]

I think the answer is 32% because if you turn 32% into a decimal (which would be 0.32) then multiply it by 25 you would get 8. Which then means if he stopped 8 times and went through 25 intersections then he stopped at 32% of the intersections. I hope this helps :)

6 0

3 years ago

Mathematically verify the outlier(s) in the data set using the 1.5 rule.

statuscvo [17]

Given:

The data values are:

7, 8, 11, 13, 14, 14, 14, 15, 16, 16, 18, 19

To find:

The outliers of the given data set.

Solution:

We have,

7, 8, 11, 13, 14, 14, 14, 15, 16, 16, 18, 19

Divide the data set in two equal parts.

(7, 8, 11, 13, 14, 14), (14, 15, 16, 16, 18, 19)

Divide each parenthesis in two equal parts.

(7, 8, 11), (13, 14, 14), (14, 15, 16), (16, 18, 19)

Now,

$Q_1=\dfrac{11+13}{2}$

$Q_1=\dfrac{24}{2}$

$Q_1=12$

And

$Q_3=\dfrac{16+16}{2}$

$Q_3=\dfrac{32}{2}$

$Q_3=16$

The interquartile range is:

$IQR=Q_3-Q_1$

$IQR=16-12$

$IQR=4$

The data values lies outside the interval $[Q_1-1.5IQR,Q_3+1.5IQR]$ are known as outliers.

$[Q_1-1.5IQR,Q_3+1.5IQR]=[12-1.5(4),16+1.5(4)]$

$[Q_1-1.5IQR,Q_3+1.5IQR]=[12-6,16+6]$

$[Q_1-1.5IQR,Q_3+1.5IQR]=[6,22]$

All the data values lie in the interval [6,22]. So, there are no outliers.

Hence, the correct option is 4.

3 0

3 years ago