Given: BE bisects AD at C, AB ⊥ BC , DE ⊥ EC, AB ≅ DE Prove: ABC ≅DEC

The following data represent the monthly phone use, in minutes, of a customer enrolled in a fraud prevention program for the pas

Illusion [34]

Answer:

The answer is "717.25 minutes".

Step-by-step explanation:

In this question first, we arrange the value in ascending order that are:

307,311,321,322,354,363,377,406,425,435,461,464,474,499,505,513,517,534,548

The median is always in an orderly position that is $= \frac{(n+1)}{2}$ .

$\to \frac{(n+1)}{2} = \frac{(20+1)}{2} = 10.5$ position orders

$10^{th} \ and \ 11^{th}$ place an average of observations so, the

$Average = \frac{(425 +435)}{2} = \frac{(860)}{2} =430$

Because as medium stands at 10.5 that median is as below 10 are greater than the level.

$Q_1$ often falls throughout the ordered BELOW average is $= \frac{(n+1)}{2}$ place.

$\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th}$ ordered position

Average $5^{th} \ and \ 6^{th}$ location findings

$Q_1 = \frac{(354+363)}{2} = \frac{(717)}{2}= 358.5$

In $= \frac{(n+1)}{2}$ the ordered place, Q3 always falls ABOVE the median.

$\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th}$ ordered At median ABOVE

Consequently $Q_3$ falls between $5^{th} \ and \ 6^{th}$ ABOVE the median position

$15^{th}\ and \ 16^{th}$ place average of observations

$\to Q_3 = \frac{(499+505)}{2}=\frac{(1004)}{2} = 502$

$\to IQR = Q_3 - Q_1= 502-358.5= 143.5$

$\to \text{Upper fence} = Q_3 + 1.5 \times IQR= 502 + 1.5 \times 143.5 = 717.25$

4 0

3 years ago

Identify the coefficient. 13 + 2q = 56

IRISSAK [1]

Answer:

2 is the coefficient

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Hey I need help pleasee!

kirill115 [55]

9514 1404 393

Answer:

Africa

Step-by-step explanation:

In order, from lowest to highest, the low points are ...

Antarctica < Asia < Africa < North America < South America < Europe < Australia

The lower continent on your answer choices list is Africa.

7 0

2 years ago

The graph shows the relationship between the hours a soccer team practiced after the season started and their total practice tim

Sergio [31]

Answer:

6 hours

Step-by-step explanation:

The picture of the question in the attached figure

Let

x ----> the hours a soccer team practiced after the season started

y ----> the total practice time for the year in hours

we know that

The y-intercept of a linear equation is the value of x when the value of y is equal to zero

In the context of this problem, the y-intercept is the total practice time before the season began (the value of x is equal to zero)

Looking at the graph

The y-intercept is the point (0,6)

therefore

the total practice time before the season began is 6 hours

5 0

3 years ago

Check my answer!

olga_2 [115]

Factor out a 6

6 / 6x^2 = x^2
6 / -12x = -2x
6 / -18 = -3

6(x^2 - 2x - 3)

Multiply the first term to the last
1 x -3 = -3

(a x b) = -3
(a + b) = -2 (middle term)

(1 x -3) = -3
(1 + -3) = -2

Plug them in by replacing -2x and simplify

6(x^2 - 1x - 3x - 3)
6((x^2 - 1x) - (3x - 3))
6(x(x - 1) - 3(x - 1))
6(x - 3)(x - 1)

Or in this case, 6(x - 1)(x - 3)

8 0

3 years ago