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gregori
19 days ago
12

A farmer estimates that he has 9,000 bees producing honey on his farm. The farmer becomes concerned when he realizes the populat

ion of bees seems to be decreasing steadily at a rate of 5% per year. If the number of bees in the population after x years is represented by f(x), which statements about the situation are true? Check all that apply. The function f(x) = 9,000(1.05)x represents the situation. The function f(x) = 9,000(0.95)x represents the situation. After 2 years, the farmer can estimate that there will be about 8,120 bees remaining. After 4 years, the farmer can estimate that there will be about 1,800 bees remaining. The domain values, in the context of the situation, are limited to whole numbers. The range values, in the context of the situation, are limited to whole numbers.
Mathematics
2 answers:
Leona [9.2K]19 days ago
5 0

The equation f(x)=9,000(0.95)x accurately depicts the situation.

After 2 years, it's estimated that there will be around 8,120 bees left.

The range values in this scenario are confined to whole numbers.

Inessa [8.9K]19 days ago
3 0
To derive the function that characterizes the bee population:

1) Initially, there are 9,000 bees in the first year.

2) In the second year, a reduction of 5% occurs => 9,000 - 0.05 * 9,000 = 9,000 * (1 - 0.05) = 9,000 * 0.95

3) Each subsequent year sees a 5% decline => 9,000 * (0.95)^(number of years)

4) Let x represent years and f(x) signify the bee count, then: f(x) = 9,000 (0.95)^x.

Evaluation of the claims:

<span>1) The function f(x) = 9,000(1.05)x applies to the scenario.

FALSE: WE ESTABLISHED IT AS f(x) = 9,000 (0.95)^x

2) The function f(x) = 9,000(0.95)x applies to the scenario.

TRUE: THIS IS THE RESULT OF OUR PRIOR ANALYSIS.

3) After 2 years, the farmer projects approximately 8,120 bees will be left.

Calculating:

f(2) = 9,000 * (0.95)^2 = 9,000 * 0.9025 = 8,122

Thus, this statement is TRUE

4) After 4 years, the farmer can predict there will be roughly 1,800 bees left.

f(4) = 9,000 * (0.95)^4 = 9,000 * 0.81450625 = 7,330

This statement is therefore FALSE

5) The domain values contextual to this situation are restricted to whole numbers.

FALSE: DOMAIN VALUES INCLUDE ALL NON-NEGATIVE REAL NUMBERS. FOR INSTANCE, THE FUNCTION IS VALID AT X = 5.5

6) The range values pertinent to this situation are restricted to whole numbers.

TRUE: FRACTIONS OF BEES CANNOT EXIST.
</span>
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The chance of winning a certain game at a carnival is 2 in 5. If Andy plays the game 12 times, what is the probability that he l
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Answer:

There is a 1.5267% chance that he loses AT MOST 3 times.

Step-by-step explanation:

Each game Andy plays has only two outcomes: a win or a loss. The probability of winning doesn’t depend on previous games, leading us to use the binomial probability distribution for this situation.

Binomial probability distribution

This distribution calculates the probability of achieving exactly x successes in n trials with only two possible outcomes for X.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

Where C_{n,x} signifies the various combinations of x objects from a set of n elements, calculated using:

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The winning probability for a certain carnival game is 2 out of 5.

This implies a losing chance of (5-2) in 5, equaling 3 out of 5.

Thus, p = \frac{3}{5} = 0.6

Considering 12 games:

Consequently, n = 12.

What probability is there that he loses AT MOST 3 times?

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

Where:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{12,0}.(0.6)^{0}.(0.4)^{12} = 0.000017

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P(X = 2) = C_{12,2}.(0.6)^{2}.(0.4)^{10} = 0.002491

P(X = 3) = C_{12,3}.(0.6)^{3}.(0.4)^{9} = 0.012457

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17 oz bottle

11 oz can

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