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We need to know three other things to answer this

1 - Area of  US  in square miles

2 - CA's  population

3 - CA's  area in square miles

The  standard deviation  = sqrt (variance)  = sqrt (4)  = 2

Triangle  BDC  is similar to triangle BEA

Therefore     EA :  DC  = 20 : 10   = 2 : 1

Therefore...there are 3 parts to the total  height of the figure

And...since the triangles are similar..it means that the height of triangle BEA  to the height of triangle BDC = 2 : 1

So....the height of  triangle  BEA  = h  =  (2/3) (36)   = 24 cm

Notice that....

area of trapezium ABDE  =  area of triangle ACE  -  area of triangle BCD

And we know that △ACE and △BCD are similar triangles, because

m∠ACE  =  m∠BCD     and     m∠BDC  =  m∠AEC  =  90°

so by the AA similarity theorem, we can say that  △ACE ~ △BCD .

Now we can find the length of CE .

\(\frac{\text{CE}}{\text{CD}}\,=\,\frac{\text{AE}}{\text{BD}}\\~\\ \frac{\text{CE}}{8\text{ cm}}\,=\,\frac{9\text{ cm}}{6\text{ cm}}\\~\\ \text{CE}\,=\,\frac{9\text{ cm}}{6\text{ cm}}\cdot8\text{ cm}\\~\\ \text{CE}\,=\,12\text{ cm} \)

Now we can find the area of  △ACE  and the area of  △BCD .

area of △ACE  =  (1/2)( CE )( AE )

area of △ACE  =  (1/2)( 12 cm )( 9 cm )

area of △ACE  =  54 cm²

area of △BCD  =  (1/2)( CD )( BD )

area of △BCD  =  (1/2)( 8 cm )( 6 cm )

area of △BCD  =  24 cm²

Now we can find the area of trapezium ABDE.

area of trapezium ABDE  =  area of △ACE  -  area of △BCD

area of trapezium ABDE  =  54 cm²  -  24 cm²

area of trapezium ABDE  =  30 cm²    

The variance is the square of the standard deviation  = 4^2   = 16

Upper exterme  = max value of data set  =  10

Five values

Minimum value = 46

Max value = 66

Median of the data set = 58  [ the red line ]  = Q₂ = Second Quartile

First quartile  = 55 = Q₁    [it's the median of the data  set points to the left of Q₂ ]

Third quartile = 63  = Q₃  [ it's the median of the data set points to the right of Q₂ ]

Example

Let   triangle ABC  have the following coordinates  ...A  = (0,0)  B  = (0, 3)  and C  = (4,0)

Let triangle DEF   have the following  coordinates ... D = (5,0)  E = (1,3)  F (1,0)

First calculate the distances between AB, AC  and BC

AB  =  sqrt  [ (0 - 0)^2 + (3 -0)^2 ] = sqrt [9] = 3

AC = sqrt [ (4 - 0)^2 + (0 -0)^2  ]  = sqrt [ 16]  = 4

BC = sqrt  [ (4 - 0)^2 + ( 3-0)^2 ]  = sqrt [ 16 +  9 ] = sqrt [25 ] = 5

Next calculate the distances between DE, DF and EF

DE  = sqrt [ (5 -1)^2 + (3 - 0)^2 ]  = sqrt [ 4^2 + 3^2 ] = sqrt [ 16 + 9 ] = sqrt [25]  = 5

DF  = sqrt [ (5 - 1)^2  + ( 0 - 0)^2 ]= sqrt [ 4^2] = sqrt [ 16] =4

EF  = sqrt [ (1 - 1)^2 + (3 - 0)^2 ] = sqrt [ 3^2]  = sqrt [9] = 3

Since AB = EF  and AC = DF  and BC = DE

Then..by S-S-S   these triangles are congruent

I'll let you see if these are congruent...remember....the distances must "match" to have S-S-S  congruecy

triangle ABC  .... A  = (0,0)   B  = (0, 4)  C  = (3,0)

triangle DEF  ...D  = ( 1,0)    E  = (4, 0)   F  = (0, 5) 

In 1997 tuition was 9,000 dollars.

In 1998 tuition was 10,000 dollars.

From 1997 to 1998 tuition increased 1,000 dollars.

In 1996 tuition was 8,500 dollars.

In 1997 tuition was 9,000 dollars.

From 1996 to 1998 tuition increased 500 dollars.

Is the increase from 1997 to 1998 twice as much as the increase from 1996 to 1998?

Is 1,000 twice as much as 500?

Yes, so it is a valid claim to say tuition increased twice as much from 1997 to 1998 as it did from 1996 to 1997.

The answer would be C as it is 2x the percentage and 60 is 1/2 of 120.

The mode is the number that occurs the most times in a list of numbers. (If two numbers occur the same number of times, and no other number occurs more times, then they are both modes.)

If we were to write down the list of ages, 15% of the numbers in the list would be 14, 50% of the numbers in the list would be 15, 20% of the numbers in the list would be 16, and 15% of the numbers in the list would be 17.

50% is the biggest percentage, so 15 is the mode.

"Oualitative" data doesn't use  "hard"  data.....it might measure "subjective" data such as  appearance, taste, etc.

So.... "B"  is qualitative  [ your tires may appear "good" to you but may appear "not good" to someone else ]

We can solve this proportion

4x / x = 48 / (x + 2)     simplify

4  = 48 / (x + 2)     multiply both sides by  (x + 2)

4(x + 2)  = 48

4x + 8 = 48     subtract 8 from both sides

4x  = 40    divide both sides by 4

x  =10

Width of the badge  = 4x  =  4 (10)   = 40 cm

the mean of a list of numbers  =  the sum of the numbers / the number of numbers

the mean of the list of medal counts  =  (93 + 101 + 110 + 103 + 121) / 5

the mean of the list of medal counts  =  528 / 5

the mean of the list of medal counts  =  105.6

Order the data from low to high

(30, 32, 33, 36)   ( 37, 40, 40, 46)

The 3rd quartile  = median of second data grouping  =  40      (1)

The 1st quartile  = median of first data grouping = [32 + 33] / 2 = 65/ 2  =  32.5     (2)

The interquartile range  =  (1) - (2) =  40 - 32.5   =   7.5  ⇒  "B"

A is incorrect...animated movies last 75 minutes...comedies, 90  minutes

B is incorrect....we only know how long each type of movie is...we don't know how many are in each category

C is incorrect....a movie  could last  4 hours!!!

D is correct.....romance movies are typically longer than comedies

Let's put each list of the yearly rainfall totals in order from least to greatest.

The yearly rainfall totals in inches for Seattle are:

23.78 ,  32.46 ,  40.30 ,  45.79 ,  49.42 ,  55.14

The median is the average of the two middle numbers in this list.

The median yearly rainfall for Seattle  =  (40.30 + 45.79)/2  in

The median yearly rainfall for Seattle  =  43.045  in

The yearly rainfall totals in inches for Las Vegas are:

2.33 ,  3.72 ,  3.84 ,  4.11 ,  5.62 ,  8.79

The median yearly rainfall for Las Vegas  =  (3.84 + 4.11)/2  in

The median yearly rainfall for Las Vegas  =  3.975  in

The difference between the median yearly rainfall for Seattle and Las Vegas   =   43.045 in  -  3.975 in

The difference between the median yearly rainfall for Seattle and Las Vegas   =   39.07 in

We can look at it this way, NSS

The total number of points he needs to average an "80"   =  5 tests * 80 average  = 400

So far....he has this many points ⇒   4 tests  *  78 average  = 312

So...he needs to make  [400 - 312] =  88  on the last test to get an average of  "80"

Mean  = [ 10 + 19 + 21 + 28 + 12 +  20 + 16 ] / 7  = 18

Squared  differences  from  mean

(10 - 18)^2  = (-8)^2  = 64

(19 - 18)^2  = 1^2  = 1

(21 - 18)^2  =  3^2  = 9

(28 -- 18)^2  = 10^2  = 100

(12 - 18)^2  = (-6)^2  = 36

(20 - 18)^2 = 2^2  = 4

(16 -  18) ^2  =  (--2)^2  = 4

Variance  = sum of  squared differences / number of values = [ 64 + 1 + 9 + 100 + 36 + 4 + 4 ] / 7  = 31.14

The range of the data set in the box and whisker plot  is  [highest value - lowest value ]  = [10 - 4  ] = 6

Arranging "A" from low to high we have  14, 20, 26, 32  ...range  = 32 - 14  = 18   

Arranging "B" from low to high we have  60, 62, 63, 63, 64 , 65  ....range  = 65 - 60  = 5

Arranging "C" from low to high  we have  16, 18, 19, 23, 26.....range = 26 - 16  = 10

Arranging "D" from low to high we have   9, 9, 11, 12, 14, 15.....range  = 15 - 9  = 6

Let x be the  missing value we need

Average (mean)  = [ sum of the values ] / [ number of values          ....so we have....

80  =  [ 71 + 91 + 82 + 90 + 88 + 61 + 70 + x ]  / 8

80  = [553  + x ]  / 8     multiply both sides by 8

640  = 553 + x        subtract 553 from both sides

87  = x  = the missing value

We need to be careful about lining up the decimals

118,000. 00

+  3,491. 04

_________

121,491.04   

11803491.04

Add 118,000.00 and 3491.04.....what do you get  ???

would it be A?