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torisob [31]
3 years ago
15

Give the domain and range of the relation.

Mathematics
1 answer:
Alex_Xolod [135]3 years ago
7 0

Answer:

Please check the explanation.

Step-by-step explanation:

Finding Domain:

We know that the domain of a function is the set of input or argument values for which the function is real and defined.

From the given graph, it is clear that the starting x-value of the line is x=-2, the closed circle at the starting value of x= -2 means the x-value x=-2 is included.

And the line ends at the x-value x=1 with a closed circle, meaning the ending value of x=1 is also included.

Thus, the domain is:

D: {-2, -1, 0, 1}       or    D: −2 ≤ x ≤ 1

Finding Range:

We also know that the range of a function is the set of values of the dependent variable for which a function is defined

From the given graph, it is clear that the starting y-value of the line is y=0, the closed circle at the starting value of y = 0 means the y-value y=0 is included.

And the line ends at the y-value y=2 with a closed circle, meaning the ending value of y=2 is also included.

Thus, the range is:

R: {0, 1, 2}   or    R: 0 ≤ y ≤ 2

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Answer: 9.9 years.

Step-by-step explanation:

If interest is compounded continuously, then formula to compute final amount A = Pe^{rt}, where P =initial amount, r= rate of interest , t=time.

Given: P= $61,000, r= 1.9% =0.019  , A = $ 73600

Substitute all values in formula

73600=61000e^{0.019t}\\\\\Rightarrow\ \dfrac{73600}{61000}=e^{0.019t}\\\\\Rightarrow\ 1.20655738=e^{0.019t}

Taking natural log on both sides

\ln (1.20655738)=0.019t\\\\\\ 0.18777116=0.019t\\\\\\ t=\dfrac{0.18777116}{0.019}\\\\\\ t=9.88269263\approx9.9\ \text{years}

Hence, the required time = 9.9 years.

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3 years ago
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Answer:

2/3

Step-by-step explanation:

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A test of abstract reasoning is given to a random sample of students before and after they completed a formal logic course. The
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Answer:

The right answer is "0.2".

Step-by-step explanation:

The given values are:

\bar{x_0} = 3.7

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As we know,

95% confidence for \mu_0 will be:

= \bar{x_0} \pm t_{\frac{0.05}{2} },n-1\times \frac{s_o}{\sqrt{n} }

The lower bound will be:

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The upper bound will be:

= 3.7+2.2621\times \frac{4.95}{\sqrt{10} }

= 7.23\simeq7.2

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