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Whitepunk [10]
2 years ago
11

Stuck on this. 60 points!

Mathematics
2 answers:
Katarina [22]2 years ago
6 0

Answer:

h(t)= (t-5)(t-11)

The swimmer comes back up 11 seconds after the timer is started.

The swimmer dives into the water 5 seconds after the timer is started.

Step-by-step explanation:

Given function:

h(t)=t^2-16t+55

To factor a quadratic in the form ax^2+bx+c,

find two numbers that multiply to ac and sum to b:

\implies ac=55

\implies b=-16

Two numbers that multiply to 55 and sum to -16 are: -11 and -5

Rewrite b as the sum of these two numbers:

\implies t^2-11t-5t+55

Factorize the first two terms and the last two terms separately:

\implies t(t-11)-5(t-11)

Factor out the common term (t - 11):

\implies (t-5)(t-11)

Therefore, the given formula in factored form is:

h(t)= (t-5)(t-11)

The swimmer's depth is modeled as h(t).  Therefore, when h(t) = 0 the swimmer will be at the surface of the water.

\implies h(t)=0

\implies (t-5)(t-11)=0

\implies t-5=0\implies t=5

\implies t-11=0 \implies t=11

Therefore, the swimmer will be at the surface of the water at 5 s and 11 s.

The swimmer's maximum depth is the vertex of the function. The x-value of the vertex is the midpoint of the zeros.  Therefore, the x-value of the vertex is t = 8.

Substitute t = 8 into the function to find the maximum depth:

\implies h(8)=8^2-16(8)+55=-9

So the swimmer's maximum depth is 9 ft.

<u>True Statements</u>

The swimmer comes back up 11 seconds after the timer is started.

The swimmer dives into the water 5 seconds after the timer is started.

Nutka1998 [239]2 years ago
4 0

Answer:

Statement 1 and 2

Step-by-step explanation:

If you take t^2 - 16t + 55 and find some of its graphical values, you will get:

Turning point: (8,-9)

Roots: (5,0) and (11,0)

When this graph is plotted and you imagine the x axis to be time (as stated in the question), each of the roots (x - intercept) must be when the swimmer goes under and when they come back up.

This means that the swimmer dived under the water at 5 seconds and came back up at 9, making the first 2 statements correct.

Now the fourth statement is ruled out.

The fifth statement is not plausible as the graph would have to have more than 2 roots for the swimmer to enter the water twice.

That leaves the third statement. If you imagine the depth of the swimmer to be the y axis of our imaginary graph, and we know that the y axis of the turning point is -9, that means that the swimmer's deepest dive was 9 feet under the water, ruling out the third statement too.

Hope this helps :D

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a) 38.4% probability that on a given day this item is requested more than 5 times.

b) 0.67% probability that on a given day this item is not requested at all.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

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In which

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e = 2.71828 is the Euler number

\mu is the mean in the given interval.

An inventory study determines that, on average, demands for a particular item at a warehouse are made 5 times per day.

This means that \mu = 5

What is the probability that on a given day this item is requested

(a) more than 5 times?

Either it is requested 5 times or less, or it is requested more than 5 times. The sum of the probabilities of these events is decimal 1. So

P(X \leq 5) + P(X > 5) = 1

We want P(X > 5). So

P(X > 5) = 1 - P(X \leq 5)

In which

P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-5}*(5)^{0}}{(0)!} = 0.0067

P(X = 1) = \frac{e^{-5}*(5)^{1}}{(1)!} = 0.0337

P(X = 2) = \frac{e^{-5}*(5)^{2}}{(2)!} = 0.0842

P(X = 3) = \frac{e^{-5}*(5)^{3}}{(3)!} = 0.1404

P(X = 4) = \frac{e^{-5}*(5)^{4}}{(4)!} = 0.1755

P(X = 5) = \frac{e^{-5}*(5)^{5}}{(5)!} = 0.1755

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P(X > 5) = 1 - P(X \leq 5) = 1 - 0.616 = 0.384

38.4% probability that on a given day this item is requested more than 5 times.

(b) not at all?

P(X = 0) = \frac{e^{-5}*(5)^{0}}{(0)!} = 0.0067

0.67% probability that on a given day this item is not requested at all.

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