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.
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Answer:
(-4,14)
(which you already choose)
Step-by-step explanation:
I wrote the domain and range in interval notation, not sure if that’s how you’re asked to do it. This is how I was taught to state the domain/range of a graph.
Answer:
Step-by-step explanation:
y > (1/3)x + 4 has an infinite number of solutions. Draw a dashed line representing y = (1/3)x + 4 and then pick points at random on either side of this line. For example, pick (1, 6). Substitute 1 for x in y > (1/3)x + 4 and 6 for y. Is the resulting inequality true? Is 6 > (1/3)(1) + 4 true? YES. So we know that (1, 6) is a solution of y > (1/3)x + 4. Because (1, 6) lies ABOVE the line y = (1/3)x + 4, we can conclude that all points abovve this line are solutions.
The other commentor is right. The correct answer is 2-i
The second follow-up question is the second option
f(x) = (x<span> – (2 + </span>i))(x<span> – (2 –</span><span> i</span>))(x<span> – 5)
</span>The third follow-up question has these three answers.
9x^2
25x
25
