1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Kryger [21]
4 years ago
6

What is the domain and range of the step function with the equation below? f(x) = -3[ x ]

Mathematics
2 answers:
Step2247 [10]4 years ago
8 0

Answer:

Option C.

Step-by-step explanation:

The given function is

f(x)=-3\left \lceil x\right \rceil

We need to find the domain and range of the given step function.

\left \lceil x\right \rceil is least integer function. The output of this function is all integers and it is defined for all values of x.

The given function is 3 times of least integer function. So, the given function f(x) is defined for all real number and outputs of the function are all integers that are multiples of 3.

Domain of f(x) = All real numbers = (-∞,∞)

Range of f(x) = All integers that are multiples of 3.

Therefore, the correct option is C.

TEA [102]4 years ago
7 0

Answer:

C

Step-by-step explanation:

The domain is all real numbers, and the range is all integers that are multiples of 3. This is right I took the test.

You might be interested in
a regular hexagonal nut of side 1 cm and length 3 cm with a central hole of diameter 1 cm. 3 cm 1 cm 1 cm. Calculate the volume
Bogdan [553]

Answer:

  5.44 cm³

Step-by-step explanation:

The volume of the hexagonal nut can be found by multiplying the area of the end face by the length of the nut. The end face area is the difference between the area of the hexagon and the area of the hole.

The area of a hexagon with side length s is given by ...

  A = (3/2)√3·s²

For s=1 cm, the area is ...

  A = (3/2)√3(1 cm)² = (3/2)√3 cm²

__

The area of a circle is given by ...

  A = πr²

The radius of a circle with diameter 1 cm is 0.5 cm. Then the area of the hole is ...

  A = π(0.5 cm)² = 0.25π cm²

__

The volume is the face area multiplied by the length, so is ...

  V = Bh = ((3/2)√3 -0.25π)(3) . . . . . cm³

  V = (9/2)√3 -0.75π cm³ ≈ 5.44 cm³

The volume of the metal is about 5.44 cm³.

5 0
3 years ago
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
The price of a sweater was reduced from $25 to &amp;19. By what percentage was the sweater reduced by?
JulijaS [17]

Answer:

24%

Step-by-step explanation:

Difference in price of sweater

=$25-$19

=$6

6/25×100%=24%

8 0
3 years ago
Mia opens a coffee shop in the first week of January. The function W(x) = 0.002x3 - 0.01x2 models the total number of customers
Len [333]
We need to know the function that models the difference in the number of customers visiting the two stores.
 We know the function that models the number of customers in the cafeteria
 W (x) = 0.002x3 - 0.01x2
 We also know the function that models the number of customers who visit the ice cream parlor 
 R (x) = x2 - 4x + 13
 Therefore the difference, D (x), in the number of customers visiting the two stores is:
 D (x) = W (x) - R (x)
 D (x) = 0.002x ^ 3 - 0.01x ^ 2 - (x ^ 2 -4x +13)
 D (x) = 0.002x ^ 3 - 0.01x ^ 2 - x ^ 2 + 4x -13

D (x) = 0.002x ^ 3 - 1.01x ^ 2 + 4x -13
<span> The answer is the third option</span>

8 0
3 years ago
Read 2 more answers
Please help me anybody.
lutik1710 [3]

Answer:

pt 1 is 3/5

Step-by-step explanation:

cuz i know

6 0
3 years ago
Read 2 more answers
Other questions:
  • I need 4 points to graph
    5·1 answer
  • A piece of copper has a mass of 58kg and displaces 5918ml. What is its density?
    9·1 answer
  • write an equation in point-slope form and slope-intercept form for a line that passes through point(4,-1) and has a slope of -3
    5·1 answer
  • A decimal number which ends after a finite numbers of a digits after the decimal point is called a blank decimal
    14·1 answer
  • the equation c = 0.75t represents c, the total cost of t tickets on a bus. which table contains values that fit this equation?
    8·1 answer
  • What is 15.8666666666666667 as a fraction
    8·1 answer
  • Expand and simplify : 3(2x+1)+2(x+3)
    5·1 answer
  • PLEASE HELP ASAP!!!! 100% CORRECT ANSWERS ONLY PLEASE!!!!
    5·2 answers
  • Plants are sold in three different sizes of tray. A small tray of 30 plants costs £6.50. A medium tray of 40 plants costs £8.95.
    5·1 answer
  • Which function is graphed?
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!