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sashaice [31]
1 year ago
7

On a coordinate plane, a curved line with a minimum value of (negative 1.25, negative 3.25) and a maximum value of (0.25, negati

ve 1.75), crosses the x-axis at (negative 2.25, 0), and crosses the y-axis at (0, negative 2). The line exits the plane at (negative 2.75, 6) and (1.5, 6).
Which statement is true about the end behavior of the graphed function?

As the x-values go to positive infinity, the function's values go to negative infinity.
As the x-values go to zero, the function's values go to positive infinity.
As the x-values go to negative infinity, the function's values are equal to zero.
As the x-values go to positive infinity, the function's values go to positive infinity.
Mathematics
1 answer:
Juliette [100K]1 year ago
8 0

The correct option is:

As the x-values go to positive infinity, the function's values go to positive infinity.

<h3>What is the end behavior of the function?</h3>

By the description, We know that the function has the points:

(-1.25, -3.25), (0.25, -1.75) (-2.25, 0), (0, -2), (-2.75, 6), (1.5, 6)

Notice that from x = 0.25 to x = 1.5 the function goes upwards, then in the right side, the function goes up, so as x goes to positive infinity, also does f(x).

On the left side, we can see  that from x = -2.75 to x = -2.25 the function goes downwards.

So, as x goes to the left (negative infinity) f(x) goes upwards.

Then as x tends to negative infinity, f(x) tends to infinity.

The correct option is:

As the x-values go to positive infinity, the function's values go to positive infinity.

If you want to learn more about end behavior:

brainly.com/question/1365136

#SPJ1

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Answer:

a.\  \mu_p=750\ \ , \sigma_p=0.005\\\\b.\  \mu_p=150\ \ , \sigma_p=0.0113\\\\c.\  \mu_p=75\ \ , \sigma_p=0.0160

Step-by-step explanation:

a. Given p=0.15.

-The mean of a sampling proportion  of n=5000 is calculated as:

\mu_p=np\\\\=0.15\times 5000\\\\=750

-The standard deviation is calculated using the formula:

\sigma_p=\sqrt{\frac{p(1-p)}{n}}\\\\=\sqrt{\frac{0.15(1-0.15)}{5000}}\\\\=0.0050

Hence, the sample mean is μ=750 and standard deviation is σ=0.0050

b. Given that p=0.15 and n=1000

#The mean of a sampling proportion  of n=1000 is calculated as:

\mu_p=np\\\\=1000\times 0.15\\\\\\=150

#-The standard deviation is calculated as follows:

\sigma_p=\sqrt{\frac{p(1-p)}{n}}\\\\\\=\sqrt{\frac{0.15\times 0.85}{1000}}\\\\\\=0.0113

Hence, the sample mean is μ=150 and standard deviation is σ=0.0113

c. For p=0.15 and n=500

#The mean is calculated as follows:

\mu_p=np\\\\\\=0.15\times 500\\\\=75

#The standard deviation of the sample proportion is calculated as:

\sigma_p=\sqrt{\frac{p(1-p)}{n}}\\\\\\=\sqrt{\frac{0.15\times 0.85}{500}}\\\\\\=0.0160

Hence, the sample mean is μ=75 and standard deviation is σ=0.0160

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