The actual distance between Emerald City and Diamond Bluff is: \(\dfrac{42}{3,5}\cdot2,75=33\left(miles\right)\)

ANSWER: 33 miles

The height of the sixth bounce is: \(5\cdot80\%\cdot80\%\cdot80\%\cdot80\%\cdot80\%=1,6384\approx1,6\left(ft\right)\)

ANSWER: 1,6 ft

Because the head size is 15% then the total size of body and foot is: 100% - 15% = 85%

The percentage of body size is: \(\dfrac{85}{\left(5+7\right)}\cdot5=\dfrac{425}{12}\approx35\%\)

P/s: I can't understand of the ratio 1:10

The boy has covered only two-fifth way => The total speed of the girl and the moving walkway is: \(\dfrac{5}{2}\)

So the ratio of the walkway's rate to the girl's rate (sililar to the boy's rate) is: \(\dfrac{5}{2}-1=\dfrac{3}{2}=3:2\)

ANSWER: \(3:2\)

The fraction is: \(\left(1-\dfrac{2}{5}\right):2=\dfrac{3}{10}\)

There are: \(\left(1992-2\right):2+1=996\left(numbers\right)\)

We arrange the numbers of dollars: 5-6-8-10-10-14-18-18-22-22-25-27-27-30-30-31-32-41-42-43-53

So the median of the dollar amount is: 25 dollars

The ratio of the area of the second and the first pasture is: \(\left(8:2\right)^2=16\)

So the area of the second pasture is: \(25\cdot16=400\left(hectares\right)\)

ANSWER: 400 hectares

There are: \(9-2=7\left(students\right)\) in orchestra and chorus

There are: \(4-2=2\left(students\right)\) in band and orchestra

So the students are in orchesta only is: \(21-\left(7+2+2\right)=10\left(students\right)\)

The ratio of triangle WXY to triangle ABC is: \(4:18=\dfrac{2}{9}\)

The base of triangle ABC is: \(243:18\cdot2=27\left(cm\right)\)

So the length of segment WY (the base of triangle WXY) is: \(27\cdot\dfrac{2}{9}=6\left(cm\right)\)

ANSWER: 6cm

Because the diagonal of the rectangle seperates the rectangle into 2 similar triangles

=> The answer is: \(\sqrt{18^2+24^2}\cdot2=\sqrt{324+576}\cdot2=\sqrt{900}\cdot2=30\cdot2=60\left(m\right)\)

We call: t-the time Tim drove

=> Tim drove at 30t mi.

=> Kim drove at: \(40\cdot3t=120t\)

=> \(t=225:\left(30+120\right)=1,5\)

So Tim drove: \(1,5\cdot30=45\left(mi\right)\)

ANSWER: 45 mi

From 101 to 199 (except from 130 to 139) she needs: \(1\cdot9=9\left(sers\right)\)

From 130 to 139 she needs: \(1\cdot10+1=11\left(sers\right)\)

From 200 to 250 (except from 230 to 239) she needs: \(1\cdot4=4\left(sers\right)\)

From 230 to 239 she needs: \(1\cdot10+1=11\left(sers\right)\)

So she need to purchase: \(11+4+11=26\left(sers\right)\)

ANSWER: 26 sers

We have: 3x + 3 = 2x + 7

=> 3x - 2x = 7 - 3

=> x = 4

So the value of x is: 4

The range of the set is: 16-4 = 12

The mean of the set is: \(\left(4+5+7+7+8+8+8+9+16\right):9=8\)

Because 12 > 8 <=> The positive difference between the range and the mean is: 12-8 = 4

The average speed which Blake traveled is: \(\dfrac{117}{2,25}=52\) (mi/h)

There are: \(1140-1140\cdot\dfrac{2}{3}=380\left(students\right)\) will not attend in the Spring Festival

He made: \(18\cdot60\%=10,8\) shots.

P/s: The number of shots mustn't be odd.

We have: \(\dfrac{a}{6}=\dfrac{b}{3}=\dfrac{c}{12}=\dfrac{a-b+c}{6-3+12}=\dfrac{30}{15}=2\)

=> \(\dfrac{a}{6}=2\) <=> \(a=6\cdot2=12\)

=> \(\dfrac{b}{3}=2\) <=> \(b=3\cdot2=6\)

=> \(\dfrac{c}{12}=2\) <=> \(c=12\cdot2=24\)

We have: \(\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{a+b}{3+4}=\dfrac{14}{7}=2\)

=> \(\dfrac{a}{3}=2\) <=> \(a=3\cdot2=6\)

=> \(\dfrac{b}{4}=2\) <=> \(b=4\cdot2=8\)