Ajug of water is pierced and begins to drip. the number of seconds between each drip is recorded. estimate the probability the next drip will occur in under 5 seconds. 1 3 2 1 1 4 6 5 6 7 7 8 9 6 9 8 6

The attached figure shows a point diagram that represents the situation. In this diagram the frequency in which the time between dripping was 1 2 3 4 5 6 7 8 and 9 seconds is observed.

we see that more often, the time between drips is 6 seconds.

However, we want to estimate the probability that the next drip occurs in less than 5 seconds.

The probability that the time between drips will be x seconds is:

P (t = x) = m / n

Where m is the number of times recorded in which the time between dripping was x seconds, and n is the total number of recorded data.

So, the probability that the time between drips is 5 seconds is:

P (t = 5) = 1/17

The probability that the time between drips is less than 5 is:

P (t <5) = P (t = 1) + P (t = 2) + P (t = 3) + P (t = 4)

P (t <5) = 3/17 + 1/17 + 1/17 + 1/17

P (t <5) = 6/17 = 0.3529

P (t <5) = 35.3%

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204