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Andre45 [30]
4 years ago
7

Iris's Botanical Garden produced flowers throughout the year. A graph demonstrating how many flowers she produced over a 11-mont

h period is shown:
A: The domain represents a 180-month period of flower production.
B: The domain represents a 11-month period of flower production.
C: The domain represents the total number of flowers produced each month.
D: The domain represents the total number of flowers produced in 11 months.

Mathematics
2 answers:
Rasek [7]4 years ago
3 0

Answer:

<em>B: The domain represents a 11-month period of flower production.</em>

Step-by-step explanation:

<u>The domain of a Function </u>

It's the set of all the values a function can take in its independent variable. The independent variable (usually x) is represented as the horizontal axis of a graph, where we can know the interval of values considered for values of the function.

In the provided graph, we can see the horizontal axis is labeled 'Months' and its values range from 1 to 11. The vertical axis is labeled "Number of flowers produced" which will be the Range of the function.

From the options presented, we can discard the C and D because they are related to the number of flowers produced, which is not the domain of the function. The option A talks about the months of production but the months range up to 180. Option B is the correct choice.

B: The domain represents a 11-month period of flower production.

asambeis [7]4 years ago
3 0

Answer:Answer:

B: The domain represents a 11-month period of flower production.

Step-by-step explanation:

The domain of a Function

It's the set of all the values a function can take in its independent variable. The independent variable (usually x) is represented as the horizontal axis of a graph, where we can know the interval of values considered for values of the function.

In the provided graph, we can see the horizontal axis is labeled 'Months' and its values range from 1 to 11. The vertical axis is labeled "Number of flowers produced" which will be the Range of the function.

From the options presented, we can discard the C and D because they are related to the number of flowers produced, which is not the domain of the function. The option A talks about the months of production but the months range up to 180. Option B is the correct choice.

B: The domain represents a 11-month period of flower production.

Step-by-step explanation: I took the test on FLVS

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Anna11 [10]
Answer :
x = 53.85

Step by step explanation:
If 78% of x (some numbers) is equal to 42, then we must set this as an equation to find it.

1) Rewriting 78% as a fraction and considering that when we say 78% of some number (x) we mean a product.
78% = 78/100 ➡️ 78/100x = 42

2) Then we can go on :
100 * 78/100x = 42 * 100
78x = 4200
78x/78 = 4200/78 ➡️ x = 53.85

3) Alternatively, simply dividing 42 by 0.78 (78%) would come to the same result :

42/0.78 = 53.85
4 0
3 years ago
Read 2 more answers
Will give brainliest to correct answer.
kirill [66]
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3 years ago
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4 0
4 years ago
A simple random sample of size n=14 is obtained from a population with μ=64 and σ=19.
madam [21]

Answer:

a) A. The population must be normally distributed

b) P(X < 68.2) = 0.7967

c) P(X  ≥  65.6) = 0.3745

Step-by-step explanation:

a) The population is normally distributed having a mean (\mu_x) = 64  and a standard deviation (\sigma_x) = \frac{19}{\sqrt{14} }

b) P(X < 68.2)

First me need to calculate the z score (z). This is given by the equation:

z=\frac{x-\mu_x}{\sigma_x} but μ=64 and σ=19 and n=14,  \mu_x=\mu=64 and \sigma_x=\frac{ \sigma}{\sqrt{n} }=\frac{19}{\sqrt{14} }

Therefore: z=\frac{68.2-64}{\frac{19}{\sqrt{14} } }=0.83

From z table, P(X < 68.2) = P(z < 0.83) = 0.7967

P(X < 68.2) = 0.7967

c) P(X  ≥  65.6)

First me need to calculate the z score (z). This is given by the equation:

z=\frac{x-\mu_x}{\sigma_x}

Therefore: z=\frac{65.6-64}{\frac{19}{\sqrt{14} } }=0.32

From z table,  P(X  ≥  65.6) =  P(z  ≥  0.32) = 1 -  P(z  <  0.32) = 1 - 0.6255 = 0.3745

P(X  ≥  65.6) = 0.3745

P(X < 68.2) = 0.7967

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3 years ago
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allsm [11]
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8 0
3 years ago
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