An ambulance travels back and forth at a constant speed along a road of length L. At a certain moment of time, an accident occurs at a point uniformly distributed on the road. [That is, the distance of the point from one of the fixed ends of the road is uniformly distributed over (0,L).] Assuming that the ambulance’s location at the moment of the accident is also uniformly distributed, and assuming independence of the variables, compute the cumulative distribution function of the distance of the ambulance from

use the power rule  

( a b ) n = a n b n

to distribute the exponent.

apply the product rule to  

3 x 2 y 3 . ( 3 x 2 ) 2 ( y 3 ) 2

apply the product rule to  

3 x 2 . 3 2 ( x 2 ) 2 ( y 3 ) 2

raise  

3

to the power of  

2 . 9 ( x 2 ) 2 ( y 3 ) 2 multiply the exponents in   ( x 2 ) 2

.

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 2 ⋅ 2 ( y 3 ) 2

multiply  

2   by   2 . 9 x 4 ( y 3 ) 2

multiply the exponents in  

( y 3 ) 2 .

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 4 y 3

⋅

2

multiply  

3   by   2 . 9 x 4 y 6

step-by-step explanation:

4th option is correct xy=-10

if this was to u plzz mark me the brainliest

the answer is 1/4

step-by-step explanation:

x= 1

step-by-step explanation: convert 5\frac{11}{12}5

​12

​

​11

​​   to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

​c

​

​b

​​ =

​c

​

​ac+b

​​

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{5\times 12+11}{12}3.25+0.5x+1

​3

​

​1

​​ +

​6

​​5

​​ x=

​12

​5×12+11

​​ 2 simplify 5\times 125×12 to 6060

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{60+11}{12}3.25+0.5x+1

​3

​​1

​​ +

​6

​5

​​ x=

​12

​​60+11

​​3 simplify 60+1160+11 to 7171

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{71}{12}3.25+0.5x+1

​3

​​1

​​ +

​6

​​5

​​ x=

​12

​​71

​​4 simplify \frac{5}{6}x

​6

​​5

​​ x to \frac{5x}{6}

​6

​5x

​​ 3.25+0.5x+1\frac{1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+1

​3

​

​1

​​ +

​6

​5x

​​ =

​12

​​71

​​ 5 convert 1\frac{1}{3}1

​3

​​1

​​   to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

​c

​​b

​​ =

​c

​​ac+b

​​3.25+0.5x+\frac{1\times 3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

​3

​​1×3+1

​​ +

​6

​​5x

​​ =

​12

​71

​​ 6 simplify 1\times 31×3 to 33

3.25+0.5x+\frac{3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

​3

​​3+1

​​ +

​6

​​5x

​​ =

​12

​71

​​ 7 simplify 3+13+1 to 44

3.25+0.5x+\frac{4}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

​3

​​4

​​ +

​6

​5x

​​ =

​12

​71

​​8 simplify 3.25+0.5x+\frac{4}{3}+\frac{5x}{6}3.25+0.5x+

​3

​​4

​​ +

​6

​​5x

​​   to \frac{13.75}{3}+\frac{4x}{3}

​3

​​13.75

​​ +

​3

​4x

​​ \frac{13.75}{3}+\frac{4x}{3}=\frac{71}{12}

​3

​​13.75

​​ +

​3

​​4x

​​ =

​12

​71

​​ 9 simplify \frac{13.75}{3}

​3

​​13.75

​​   to 4.5833334.583333

4.583333+\frac{4x}{3}=\frac{71}{12}4.583333+

​3

​​4x

​​ =

​12

​71

​​ 10 subtract 4.5833334.583333 from both sides

\frac{4x}{3}=\frac{71}{12}-4.583333

​3

​

​4x

​​ =

​12

​

​71

​​ −4.583333

11 simplify \frac{71}{12}-4.583333

​12

​

​71

​​ −4.583333 to \frac{4}{3}

​3

​4

​\frac{4x}{3}=\frac{4}{3}

​3

​​4x

​​ =

​3

​4

​​ 12 multiply both sides by 33

4x=\frac{4}{3}\times 34x=

​3

​4

​​ ×3

13 simplify \frac{4}{3}\times 3

​3

​​4

​​ ×3 to \frac{12}{3}

​3

​​12

​​ 4x=\frac{12}{3}4x=

​3

​​12

​​ 14 simplify \frac{12}{3}

​3

​​12

​​   to 44

4x=44x=4

15 divide both sides by 44

x=1

​​