<span>given:
increase in inventories= $13
increase in accounts receivable =$29
increase in accounts payable=$17
solution:
change in net working capital = increase in inventories + increase in accounts receivable - increase in accounts payable.
change in net working capital = 13+29-17 = $25</span>
You can do this by drawing one line through parallel to PQS to meet RQ at T
Now calculate length of RT:-
cos 70 = RT / 70 giving RT = 23.94m
sin 70 = ST/70 giving ST = 65.78 m
draw a line from S perpendicular to PQ to meet PQ at U.
PU = 110 - 65.78 = 44.22 m
tan 50 = SU / 44.22 giving SU = 52.70 m
TQ = SU = 52.70 m
So x = TQ + RT = 52.70 + 23.94 = 76.6 m to 1 dec place.
Answer:
Step-by-step explanation:
y = -3(x-3)^2 + 4
Answer:
$2.7
Step-by-step explanation:
Let the price of one apple be x
and the price of one orange y
Hence ;
One apple and 3 oranges would cost:
x + 3×y = $5.10----------------------------(1)
One apple and 5 oranges would cost;
x + 5×y = $7.50--------------------------(2)
Subtracting eqn (1) from (2), we have :
x-x + 5y -3y = 7.5- 5.1
2y = 2.4
y = $1.2
From eqn(1)
x + 3y = $5.10
x= $5.10-3($1.2) = $5.10-$3.6 = $1.5
Hence x=$1.5 and y= $1.2
We are required to find the cost of one apple and one orange, hence :
x+y = $1.5+$1.2= $2.7
Answer:
a) 0.1353
b) 0.3679
Step-by-step explanation:
Let's start by defining the random variable T.
T : ''The time (in hours) required to repair a machine''
T ~ exp (λ)
T ~ exp (1)
The probability density function for the exponential distribution is
(In the equation I replaced λ = L)

With L > 0 and x ≥ 0
In this exercise λ = 1 ⇒

For a)





For b)

The event (T ≥ 10 / T > 9) is equivalent to the event T ≥ 1 so they have the same probability of occur


