If you were solving it as an equation the width would be 21.
The number may be written as
Answer: 7.2008 x 10⁵
Answer:
S= 5, L= 12
Step-by-step explanation:
To find the value of L and S, we need to first label the 2 equations.
L +S= 17 -----(1)
3L +2S= 46 -----(2)
From (1):
L= 17 -S -----(3) <em>(</em><em>-S</em><em> </em><em>on</em><em> </em><em>both</em><em> </em><em>sides</em><em>)</em>
Subst. (3) into (2):
3(17-S) +2S= 46
3(17) -3S +2S= 46 <em> </em><em>(</em><em>expand</em><em>)</em>
51 -S= 46
-S= 46 -51 <em>(</em><em>-51</em><em> </em><em>on</em><em> </em><em>both</em><em> </em><em>sides</em><em>)</em>
-S= -5
S= 5 <em>(</em><em>divide</em><em> </em><em>by</em><em> </em><em>-1</em><em> </em><em>throughout</em><em>)</em>
Subst. S=5 into (3):
L= 17 -5
L= 12
Answer:
(a) $54
(b) $60
(c) $30
(d) $6.3
Step-by-step explanation:
Suppose a month has 30 days. Then the expected expense for the whole month is 30 times the expected expense at each day
(a) E[F] = 30*30%*$6 = 30*0.3*6 = $54
(b) E[T] = 30*50%*$4 = 30*0.5*4 = $60
(c) E[C] = 30*4%*$25 = 30*0.04*25 = $30
(d) Suppose all events are independent, the probability that she would spend on all 3 items on the same day is
P[F+T+C] = 0.3*0.5*0.04 = 0.006
So the expected expense she would make on all 3 items is
E[F+T+C] = 30*0.006*(6+4+25) = $6.3
No answer possible as you didn’t supply the diagram