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marissa [1.9K]
2 years ago
8

The table shows the estimated number of bees, y, in a hive x days after a pesticide is released near the hive.

Mathematics
2 answers:
Oksana_A [137]2 years ago
6 0

Answer:

<h2>A.</h2>

Step-by-step explanation:

The given table is

Days     Bees

  0        10,000

 10        7,500

 20       5,600

 30       4,200

 40       3,200

 50       2,400

Where x represents days and y represents bees.

The exponential function that models this problem must be like

y=a(1-r)^{x}, which represenst an exponential decary, because in this case, the number of bees decays.

We nned to use one points, to find the rate of decay. We know that a=10,000, because it starts with 10,000 bees.

Let's use the points (10, 7500)

y=a(1-r)^{x}\\7500=10000(1-r)^{10}

Solving for r, we have

\frac{7500}{10000}=(1-r)^{10} \\(1-r)^{10} =0.75

Using logarithms, we have

ln((1-r)^{10}) =ln(0.75)\\10 \times ln(1-r)=ln(0.75)\\ln(1-r)=\frac{ln(0.75)}{10} \approx -0.03\\e^{ln(1-r)}=e^{-0.03}\\1-r =e^{-0.03}\\r=-e^{-0.03}+1 \approx 1.97

Replacing all values in the model, we have

y=10000(1-1.97)^{x}\\y=10000(0.97)^{x}

Therefore, the right answer is the first choice, that's the best approximation to this situation.

Yanka [14]2 years ago
4 0

Answer:

A. y = 9,958(0.972)x

Step-by-step explanation:

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A small regional carrier accepted 19 reservations for a particular flight with 17 seats. 14 reservations went to regular custome
Ludmilka [50]

Answer:

(a) The probability of overbooking is 0.2135.

(b) The probability that the flight has empty seats is 0.4625.

Step-by-step explanation:

Let the random variable <em>X</em> represent the number of passengers showing up for the flight.

It is provided that a small regional carrier accepted 19 reservations for a particular flight with 17 seats.

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The random variable <em>X</em> thus follows a Binomial distribution with parameters <em>n</em> = 5 and <em>p</em> = 0.52.

(1)

Compute the probability of overbooking  as follows:

P (Overbooking occurs) = P(More than 3 shows up for the flight)

                                        =P(X>3)\\\\={5\choose 4}(0.52)^{4}(1-0.52)^{5-4}+{5\choose 5}(0.52)^{5}(1-0.52)^{5-5}\\\\=0.175478784+0.0380204032\\\\=0.2134991872\\\\\approx 0.2135

Thus, the probability of overbooking is 0.2135.

(2)

Compute the probability that the flight has empty seats as follows:

P (The flight has empty seats) = P (Less than 3 shows up for the flight)

=P(X

Thus, the probability that the flight has empty seats is 0.4625.

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