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Butoxors [25]
3 years ago
13

Suppose f(x)=x+2. Find the graph of f(1/3x)

Mathematics
1 answer:
8090 [49]3 years ago
6 0

Answer:

f(x)=x+2.

f(1/3x)=1/3 x+2=(x+6)/3or x/3+2

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Slope =1.000/2.000=0.5 X-intercept=10/-1=-10.0 Y-intercept=10/2=5
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For the following exercises, determine whether the relation is a function.
elixir [45]

Answer:

The given relation $\{(a, b),(c, d),(e, d)\}$ is a function.

Step-by-step explanation:

A relation $\{(a, b),(c, d),(e, d)\}$ is given.

It is required to determine whether the given relation is a function.

To determine whether the given function is a relation, identify the domain and range and then check whether the given relation is a function.

Step 1 of 1

The given relation is {(a, b),(c, d),(e, d)}.

The set of the first components of each ordered pair is called the domain.

From the relation, the domain is {a, c, e}.

The set of the second components of each ordered the range.

From the relation, the range is {b, d, d}.

The given relation is a function. But it is not a one-to-one function.

7 0
2 years ago
The first term of an arithmetic sequence is -5, and the tenth term is 13. Find the common difference.
dlinn [17]

\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\[-0.5em] \hrulefill\\ a_1=-5\\ n=10\\ a_{10}=13 \end{cases} \\\\\\ a_{10}=a_1+(10-1)d\implies 13=-5+(10-1)d \\\\\\ 13=-5+9d\implies 18=9d\implies \cfrac{18}{9}=d\implies \boxed{2=d}

7 0
3 years ago
A recent study from the University of Virginia looked at the effectiveness of an online sleep therapy program in treating insomn
Ludmilka [50]

Answer:

1. The 99% confidence interval for the difference in average is -6.47377 < μ₁ - μ₂ < -11.34623

2. The possible issues in the calculations includes;

a. The confidence level used in the confidence interval can influence the result of the confidence interval observed

b. The sample size is small

Step-by-step explanation:

1. The number of adults with insomnia in the sample = 45

The number of adults that participated in the therapy, n₁ = 22

The number of candidates that served as control group, n₂ = 23

The average score for the for the 22 participants of the program, \overline x_1 = 6.59

The standard deviation for the 22 participants of the program, s₁ = 4.10

The average score for the for the 23 subjects in the control group, \overline x_2 = 15.50

The standard deviation for the 23 subjects in the control group, s₂ = 5.34

The confidence interval for unknown standard deviation, σ, is given by the following expression;

\left (\bar{x}_{1}- \bar{x}_{2}  \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}

α = 1 - 0.99 = 0.01

α/2 = 0.005

The degrees of freedom, df = 22 - 1 = 21

t_{\alpha /2} = t_{0.005, \, 21} = 1.721

Therefore, we have;

\left (6.59- 15.5  \right )\pm1.721 \cdot \sqrt{\dfrac{4.10^{2}}{22}+\dfrac{5.34^{2}}{23}}

The 99% confidence interval for the difference in average is therefore given as follows;

-6.47377 < μ₁ - μ₂ < -11.34623

Therefore, there is considerable evidence that the participants in the survey  had lower average score than the subjects in the control group

2. The possible issues in the calculations are;

a. The confidence level used in the confidence interval can influence the result of the confidence interval observed

b. The sample size is small

5 0
2 years ago
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