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Kitty [74]
3 years ago
5

What is the domain of f(x) = 5- 7?

Mathematics
1 answer:
anyanavicka [17]3 years ago
6 0

Answer:

c

Step-by-step explanation:

hi

You might be interested in
The principal at Crest Middle School, which enrolls only sixth-grade students and seventh-grade students, is interested in deter
AlekseyPX

Answer:

a) [ -27.208 , -12.192 ]

b) New procedure is not recommended

Step-by-step explanation:

Solution:-

- It is much more common for a statistical analyst to be interested in the difference between means than in the specific values of the means themselves.

- The principal at Crest Middle School collects data on how much time students at that school spend on homework each night.  

- He/She takes a " random " sample of n = 20 from a sixth and seventh grades students from the school population to conduct a statistical analysis.

- The summary of sample mean ( x1 & x2 ) and sample standard deviation ( s1 & s2 ) of the amount of time spent on homework each night (in minutes) for each grade of students is given below:

                                                          <u>Mean ( xi )</u>       <u> Standard deviation ( si )</u>

          Sixth grade students                 27.3                            10.8                  

          Seventh grade students           47.0                             12.4

- We will first check the normality of sample distributions.

  • We see that sample are "randomly" selected.
  • The mean times are independent for each group
  • The groups are selected independent " sixth " and " seventh" grades.
  • The means of both groups are conforms to 10% condition of normality.

Hence, we will assume that the samples are normally distributed.

- We are to construct a 95% confidence interval for the difference in means ( u1 and u2 ).

- Under the assumption of normality we have the following assumptions for difference in mean of independent populations:

  • Population mean of 6th grade ( u1 ) ≈ sample mean of 6th grade ( x1 )  
  • Population mean of 7th grade ( u2 ) ≈ sample mean of 6th grade ( x2 )

Therefore, the difference in population mean has the following mean ( u ^ ):

                      u^ = u1 - u2 = x1 - x2

                      u^ = 27.3 - 47.0

                      u^ = -19.7

- Similarly, we will estimate the standard deviation (Standard Error) for a population ( σ^ ) represented by difference in mean. The appropriate relation for point estimation of standard deviation of difference in means is given below:

                    σ^ =  √ [ ( σ1 ^2 / n1 ) + ( σ2 ^2 / n2 ) ]

Where,

          σ1 ^2 : The population variance for sixth grade student.

          σ2 ^2 : The population variance for sixth grade student.

          n1 = n2 = n : The sample size taken from both populations.

Therefore,

                 σ^ =  √ [ ( 2*σ1 ^2   / n )].

- Here we will assume equal population variances : σ1 ≈ σ2 ≈ σ is "unknown". We can reasonably assume the variation in students in general for the different grade remains somewhat constant owing to other reasons and the same pattern is observed across.

- The estimated standard deviation ( σ^ ) of difference in means is given by:

σ^ =

           s_p*\sqrt{\frac{1}{n_1} + \frac{1}{n_2}  } = s_p*\sqrt{\frac{1}{n} + \frac{1}{n}  } = s_p*\sqrt{\frac{2}{n}}\\\\\\s_p = \sqrt{\frac{(n_1 - 1 )*s_1^2 + (n_2 - 1 )*s_2^2}{n_1+n_2-2} } =  \sqrt{\frac{(n - 1 )*s_1^2 + (n - 1 )*s_2^2}{n+n-2} } = \sqrt{\frac{(n - 1 )*s_1^2 + (n - 1 )*s_2^2}{2n-2} } \\\\s_p = \sqrt{\frac{(20 - 1 )*s_1^2 + (20 - 1 )*s_2^2}{2(20)-2} } \\\\s_p = \sqrt{\frac{19*10.8^2 + 19*12.4^2}{38} } = \sqrt{135.2}  \\\\s_p = 11.62755

           σ^ = 11.62755*√2/20

          σ^ = 3.67695

- Now we will determine the critical value associated with Confidence interval ( CI ) which is defined by the standard probability of significance level ( α ). Such that:

         Significance Level ( α ) = 1 - CI = 1 - 0.95 = 0.05

                   

- The reasonable distribution ( T or Z ) would be determined on the basis of following conditions:

  • The population variances ( σ1 ≈ σ2 ≈ σ )  are unknown.
  • The sample sizes ( n1 & n2 ) are < 30.

Hence, the above two conditions specify the use of T distribution critical value. The degree of freedom ( v ) for the given statistics is given by:

          v = n1 + n2 - 2 = 2n - 2 = 2*20 - 2

          v = 38 degrees of freedom        

- The t-critical value is defined by the half of significance level ( α / 2 ) and degree of freedom ( v ) as follows:

          t-critical = t_α / 2, v = t_0.025,38 = 2.024

- Then construct the interval for 95% confidence as follows:

          [ u^ - t-critical*σ^ , u^ + t-critical*σ^ ]

          [ -19.7 - 2.042*3.67695 , -19.7 + 2.042*3.67695 ]

          [ -19.7 - 7.5083319 , -19.7 + 7.5083319 ]

          [ -27.208 , -12.192 ]

- The principal should be 95% confident that the difference in mean times spent of homework for ALL 6th and 7th grade students in this school (population) lies between: [ -27.208 , -12.192 ]

- The procedure that the matched-pairs confidence interval for the mean difference in time spent on homework prescribes the integration of time across different sample groups.

- If we integrate the times of students of different grades we would have to  make further assumptions like:

  • The intelligence levels of different grade students are same
  • The aptitude of students from different grades are the same
  • The efficiency of different grades are the same.

- We have to see that both samples are inherently different and must be treated as separate independent groups. Therefore, the above added assumptions are not justified to be used for the given statistics. The procedure would be more bias; hence, not recommended.

                 

8 0
3 years ago
Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x.
charle [14.2K]

ANSWER

See below

EXPLANATION

Given

f(x) =  \frac{ {x}- 9 }{x + 5}

and

g(x) =   \frac{ - 5x - 9}{x - 1}

(f \circ \: g)(x)=  \frac{ (\frac{ - 5x - 9}{x - 1})- 9 }{(\frac{ - 5x - 9}{x - 1} )+ 5}

(f \circ \: g)(x)=  \frac{ \frac{ - 5x - 9 - 9(x - 1)}{x - 1}}{\frac{ - 5x - 9 + 5(x - 1)}{x - 1} }

Expand:

(f \circ \: g)(x)=  \frac{ \frac{ - 5x - 9 - 9x  + 9}{x - 1}}{\frac{ - 5x - 9 + 5x - 5}{x - 1} }

(f \circ \: g)(x)=  \frac{ \frac{ - 5x - 9x  + 9 - 9}{x - 1}}{\frac{ - 5x + 5x - 5 - 9}{x - 1} }

(f \circ \: g)(x)=  \frac{ \frac{ - 14x }{x - 1}}{\frac{ -14}{x - 1} }

Since the denominators are the same, they will cancel out,

(f \circ \: g)(x)= \frac{ - 14x}{ - 14}  = x

8 0
2 years ago
Can anyone find me a chapter 10 extra credit for algebra 1
inysia [295]

Answer:

i think i found it, but you might have to print it out, but i will send the link: https://www.murrieta.k12.ca.us/cms/lib5/CA01000508/Centricity/Domain/1791/_Files/Algebra%201/c10rev.pdf

:) have a nice day love ❤

Step-by-step explanation:

7 0
3 years ago
Find the sum (6x2+8x+8)+(12x3-12x2-2x)
Maslowich
 The sum is 12x3+18x2-10x+8
7 0
3 years ago
PLEASEE HELP!
melamori03 [73]

Answer:

Two non zero vectors, a and b are parallel when they are scalar multiples of each other such that a = c·b where c is a scalar quantity.

Therefore, in order to find a vector that is parallel to the vector, b = (-2, -1), we multiply the vector, b by a scaler quantity

Step-by-step explanation:

Given that the vector b = (-2, -1) can be written as follows;

b = -2·i - j, we have;

\left | b \right | = √((-2)² + (-1)²) = √5

Therefore, we have;

The coordinates of the endpoint of the vector are (-2, 0) and (0, -1)

Therefore, the slope of the vector = (-1 - 0)/(0 - (-2)) = -1/2

The slope of parallel vectors are equal, which gives the slope of the parallel vector = -1/2 = (λ × (-1 - 0))/(λ ×(0 - (-2))

Therefore, a parallel vector is obtained from a vector by multiplying with a scaler product.

7 0
2 years ago
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