Triangle ABC is transformed to triangle A′B′C′, as shown below:

A number r is no less than −1.5 and fewer than 9.5

melisa1 [442]

Answer:

r can be any number -1.5 to 9.4999999

Step-by-step explanation:

4 0

3 years ago

Please help. I don't understand anything.

alina1380 [7]

Answer:

A

Step-by-step explanation:

The area of figure=Area of rectangle+Area of triangle

Area of figure=(x+2)(x-3)+1/2*(x)*(x+2)

Area of figure=x^2-x-6+0.5*x^2+x

Area of figure=(3/2)x^2-6

7 0

2 years ago

Read 2 more answers

Type the correct answer in each box. Use numerals instead of words. Any non-integer answers in this problem should be entered as

ladessa [460]

Answer:

The interquartile range of the data set is 5

The mean absolute deviation of the data set is 3.6

Step-by-step explanation:

* Lets explain how to find the interquartile range and the mean absolute

deviation (MAD)

- The steps to find the interquartile range is:

1- Arrange the values from the smallest to the largest

∴ The values are 5 , 6 , 10 , 11 , 12 , 13 , 14 , 15 , 18 , 20

2- Find the median

- The median is the middle value after arrange them

* If there are two values in the middle take their average

∵ The values are 10 then the 5th and the 6th are the values

∵ The 5th is 12 and the 6th is 13

∴ The median = $\frac{12+13}{2}=12.5$

∴ The median is 12.5

3- Calculate the median of the lower quartile

- The lower quartile is the median of the first half data values

∵ There are 10 values

∴ The first half is the first five values

∴ The first half values are 5 , 6 , 10 , 11 , 12

∵ The middle value is 10

∴ The median of lower quartile = 10

- Similar find the median of the upper quartile

- The upper quartile is the median of the second half data values

∵ There are 10 numbers

∴ The second half is the last five values

∴ The second half values are 13 , 14 , 15 , 18 , 20

∵ The middle value is 15

∴ The median of upper quartile = 15

4- The interquartile range (IQR) is the difference between the upper

and the lower medians

∴ The interquartile range = 15 - 10 = 5

* The interquartile range of the data set is 5

* Lets talk about the mean absolute deviation

- Mean absolute deviation (MAD) of a data set is the average distance

between each data value and the mean

- To find the mean absolute deviation of the data, start by finding

the mean of the data set.

1- Find the sum of the data values, and divide the sum by the

number of data values.

∵ The data set is 5 , 6 , 10 , 11 , 12 , 13 , 14 , 15 , 18 , 20

∵ Its sum = 5 + 6 + 10 + 11 + 12 + 13 + 14 + 15 + 18 + 20 = 124

∵ The mean = the sum of the data values/the number of the data

∵ The set has 10 numbers

∴ The mean = 124/10 = 12.4

2- Find the absolute value of the difference between each data value

and the mean ⇒ |data value – mean|

# I5 - 12.4I = 7.4

# I6 - 12.4I = 6.4

# I10 - 12.4I = 2.4

# I11 - 12.4I = 1.4

# I12 - 12.4I = 0.4

# I13 - 12.4I = 0.6

# I14 - 12.4I = 1.6

# I15 - 12.4I = 2.6

# I18 - 12.4I = 5.6

# I20 - 12.4I = 7.6

3- Find the sum of the absolute values of the differences.

∵ Their sum = 7.4 + 6.4 + 2.4 + 1.4 + 0.4 + 0.6 + 1.6 + 2.6 + 5.6 + 7.6

∴ Their sum = 36

4- Divide the sum of the absolute values of the differences by the

number of data values to find MAD

∴ MAD = The sum of the absolute values/number of the values

∵ The sum = 36

∵ The data set has 10 values

∴ MAD = 36/10 = 3.6

* The mean absolute deviation of the data set is 3.6

6 0

2 years ago

The fish population of Lake Collins is decreasing at a rate of 5% per year. In 2004 there were about 1,350 fish. Write an expone

ad-work [718]

Answer:

Part 1) The exponential function is equal to $y=1,350(0.95)^{x}$

Part 2) The population in 2010 was $992\ fish$

Step-by-step explanation:

Part 1) Write an exponential decay function that models this situation

we know that

In this problem we have a exponential function of the form

$y=a(b)^{x}$

where

y ----> the fish population of Lake Collins since 2004

x ----> the time in years

a is the initial value

b is the base

we have

$a=1,350\ fish$

$b=(100\%-5\%)=95\%=0.95$

substitute

$y=1,350(0.95)^{x}$ ----> exponential function that represent this scenario

Part 2) Find the population in 2010

we have

$y=1,350(0.95)^{x}$

so

For $x=(2010-2004)=6\ years$

substitute

$y=1,350(0.95)^{6}=992\ fish$

7 0

3 years ago

Convert the following equations from point-slope form to slope-intercept form.

mojhsa [17]

<h2>Answer:</h2>

We need to solve both for y.

7 0

3 years ago