help (1 Viewer)

[SKA]

so speed = distance over time

how do u do this quesiton :

The stopping distance of a car is proportional to the square of the car's speed. A car travelling at 60 km/h has a stopping distance of 40m.

If the stopping distance is 80m, what is the car's speed?

 

[SKA]

i struggle with probablitiy.. how do u do this :'(

James has six different-coloured pencils in his pencil case. He takes out two pencils without looking at their colours.

How many different combinations of colours are possible?

 

[aim54x]

i struggle with probablitiy.. how do u do this :'(

James has six different-coloured pencils in his pencil case. He takes out two pencils without looking at their colours.

How many different combinations of colours are possible?

use ur calculator type in 6C2.

 

[aim54x]

the other question i havn't come across b4 i should b able to work it out though. may get back 2 u on that.

 

[SKA]

Sarah has two packets of jelly beans. Each packet contains one black and five yellow jelly beans. Sarah takes one jelly bean from each packet without looking.

What is the probability that both of the jelly beans are black?



stupid probability... answer says 1/36????

 

[SKA]

use ur calculator type in 6C2.

i forgot about the calculator way

 

[Frith]

there is 1 black jelly bean in each packet so thats 1 over 6 and there are two packets so its just 1/6 x 1/6 which should get you 1/36.(its one 1/6 beacuase we are looking for the prob of the black beans)

i dunno about the speed one is it the square root of 7200?just an educated guess

 

[malkin86]

so if she pulled out two jelly beans and replaced them 36 times, she'd get one time (in the world of probability) where she got both black jellybeans.

 

[malkin86]

so speed = distance over time

how do u do this quesiton :

The stopping distance of a car is proportional to the square of the car's speed. A car travelling at 60 km/h has a stopping distance of 40m.

If the stopping distance is 80m, what is the car's speed?

I'll give this a burl...

X = kY

60km (squared) = k.40m

3600 / 0.40 = k

k = 9000

substitute back in and try again for 80m...

9000 x 80 = speedsquared

= 720000

speed = squareroot ans

speed = 848.5 km/h...

hey... that can't be right! Oh well, I hope it jogs someone's memory...

 

[lilmzqt]

so speed = distance over time

how do u do this quesiton :

The stopping distance of a car is proportional to the square of the car's speed. A car travelling at 60 km/h has a stopping distance of 40m.

If the stopping distance is 80m, what is the car's speed?

nothing to do with speed = distance over time.

try this(D=stopping distance, S=speed)

D=kS (s is squared)
60=k40 (40 needa b sqaured)

solve that to get wat k is. then to find S wen the stopping distance is 80 put 80 in tha formula D=kS (squared) and put tha value of k instead of k in and solve it as an equation.

 

[SKA]

nothing to do with speed = distance over time.

try this(D=stopping distance, S=speed)

D=kS (s is squared)
60=k40 (40 needa b sqaured)

solve that to get wat k is. then to find S wen the stopping distance is 80 put 80 in tha formula D=kS (squared) and put tha value of k instead of k in and solve it as an equation.

oh dat seems much easier..
thanks