Answer: $74
<u>Step-by-step explanation:</u>
Let C represent the cost of a coach ticket
Let B represent the cost of a business ticket
Washington: 150C + 50B = 18,450
NY City: 170C + 100B = 27,280
Solve the system using the Elimination method:
-2(150C + 50B = 18,450) → -300C - 100B = -36,900
1(170C + 100B = 27,280) → <u>170C + 100B = 27,280 </u>
-130C = - 9,620
<u> ÷ -130 </u> <u> ÷ -130 </u>
C = 74
8x-12y=84
in slope intersept form or (y=mx+b) is
y=2/3x-7
If the probability of observing at least one car on a highway during any 20-minute time interval is 609/625, then the probability of observing at least one car during any 5-minute time interval is 609/2500
Given The probability of observing at least one car on a highway during any 20 minute time interval is 609/625.
We have to find the probability of observing at least one car during any 5 minute time interval.
Probability is the likeliness of happening an event among all the events possible. It is calculated as number/ total number. Its value lies between 0 and 1.
Probability during 20 minutes interval=609/625
Probability during 1 minute interval=609/625*20
=609/12500
Probability during 5 minute interval=(609/12500)*5
=609/2500
Hence the probability of observing at least one car during any 5 minute time interval is 609/2500.
Learn more about probability at brainly.com/question/24756209
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Answer:
166,833
Step-by-step explanation:
t
n
=
a
+
(
n
−
1
)
d
999
=
3
+
(
n
−
1
)
3
999
=
3
+
3
n
−
3
999
=
3
n
n
=
333
We can now use the formula
s
n
=
n
2
(
2
a
+
(
n
−
1
)
d
)
)
to determine the sum.
s
333
=
333
2
(
2
(
3
)
+
(
333
−
1
)
3
)
s
333
=
333
2
(
1002
)
s
333
=
166
,
833
Hopefully this helps!
