Order doesn't mater so use combinations:
6C3 = 6!/ (3! (6-3)! ) = 6! /(3!*3!) = 6*5*4*3! / (3! 3!) = 6*5*4/3*2*1) = 20
Answer:
6 cm
Step-by-step explanation:
Vcuboid = lwh
⇒ h = Vcuboid ÷ lw
∴ h = 540 cm³ ÷ 90 cm²
⇒ h = 6 cm
Answer: There are 5 grasshoppers, 15 crickets and 20 ladybugs
Step-by-step explanation:
The question in english is:
Jose has found several insects. There are 10 less grasshoppers than crickets. There are 5 less crickets than ladybugs. If Jose has found 5 grasshoppers, how many ladybugs did he find and how many crickets? Write two equations to solve the problem.
Let's tag grasshoppers with 
, crickets with 
and ladybugs with .
So, we are tolde there are 10 less grasshoppers than crickets:
(1) We have the first equation
Then, we are told there are 5 less crickets than ladybugs:
(2) This is the second equation
If 
and we substitute this in both equations we will have:
(3)
Isolating :
(4) There are 15 crickets
Substituting (4) in (2):
(5)
Isolating 
:
There are 20 ladybugs
Therefore:
There are 15 crickets and 20 ladybugs
Answer:
Step-by-step explanation:
Hope this helps.
<em>a</em> = 8 - a train arrives at the station once every 8 minutes, so for any given 8 minute interval, a randomly selected train has uniform probability of arriving at the station at some point in this time.
<em>f(x)</em> = 1/8 - the area under the graph of <em>f(x)</em> must be equal to 1. This area corresponds to a rectangle with length <em>a</em> = 8 and height <em>x</em> such that 8<em>x</em> = 1. Solving for <em>x</em> gives 1/8.
<em>P</em> = 5/8 - this is equal to the area under the graph over the interval [0, 5], which is the area of a rectangle with length 5 and height 1/8.
