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Veseljchak [2.6K]
3 years ago
8

each statement below describes the transformation of the graph f(x)=x square. which statement correctly describes the graph of g

(x)=(x-7)^2 +7

Mathematics
1 answer:
Vitek1552 [10]3 years ago
8 0

Answer:

The correct option is:

The graph of g(x) is the graph of f(x) translated 7 units up and 7 units right.

Option A is correct option.

Step-by-step explanation:

The parent function is: f(x)=\sqrt{x}

The transformed function is: g(x)=\sqrt{x-7}  +7

We need to find the statement that best describes the transformed function.

We know the transformation rule:

If f(x) is transformed into f(x)+c, then the function is transformed vertically c units up.

If f(x) is transformed into f(x)-c, then the function is transformed vertically c units down.

If f(x) is transformed into f(x-c) then the function is transformed right c units.

If f(x) is transformed into f(x+c) then the function is transformed left c units.

So, In the given transformation:

The parent function is: f(x)=x^2

The transformed function is: g(x)=\sqrt{x-7}  +7

The transformed function is shifted 7 units up g(x)=\sqrt{x-7} \mathbf{ +7} and 7 units right g(x)=\sqrt{x\mathbf{-7}} +7

So, The correct option is:

The graph of g(x) is the graph of f(x) translated 7 units up and 7 units right.

Option A is correct option.

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2 years ago
Assuming that x=-2 and y=3 evaluate xy-x^3 ​
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(-2^3) = -8.
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2 years ago
Can some please help me with this math question.
Veseljchak [2.6K]

Answer:

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6 0
3 years ago
Construct a​ 99% confidence interval for the population​ mean, mu. Assume the population has a normal distribution. A group of 1
Zarrin [17]

Answer:

99% confidence interval for the population​ mean is [19.891 , 24.909].

Step-by-step explanation:

We are given that a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.

Assuming the population has a normal distribution.

Firstly, the pivotal quantity for 99% confidence interval for the population​ mean is given by;

         P.Q. = \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, \bar X = sample mean age of selected students = 22.4 years

             s = sample standard deviation = 3.8 years

             n = sample of students = 19

             \mu = population mean

<em>Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.</em>

So, 99% confidence interval for the population​ mean, \mu is ;

P(-2.878 < t_1_8 < 2.878) = 0.99  {As the critical value of t at 18 degree of

                                                freedom are -2.878 & 2.878 with P = 0.5%}

P(-2.878 < \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } < 2.878) = 0.99

P( -2.878 \times {\frac{s}{\sqrt{n} } } < {\bar X - \mu} < 2.878 \times {\frac{s}{\sqrt{n} } } ) = 0.99

P( \bar X -2.878 \times {\frac{s}{\sqrt{n} } < \mu < \bar X +2.878 \times {\frac{s}{\sqrt{n} } ) = 0.99

<u>99% confidence interval for</u> \mu = [ \bar X -2.878 \times {\frac{s}{\sqrt{n} } , \bar X +2.878 \times {\frac{s}{\sqrt{n} } ]

                                                 = [ 22.4 -2.878 \times {\frac{3.8}{\sqrt{19} } , 22.4 +2.878 \times {\frac{3.8}{\sqrt{19} } ]

                                                 = [19.891 , 24.909]

Therefore, 99% confidence interval for the population​ mean is [19.891 , 24.909].

6 0
3 years ago
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