Review:

Which of the following would be a correct interpretation of a 99% confidence interval such as 4.1 less than muless than5.6? Ch

Harrizon [31]

Answer: A. We are 99% confident that the interval from 4.1 to 5.6 actually does contain the true value of mu.

Step-by-step explanation:

The correct interpretation of a 99% confidence interval is that we are 99% confident that the true population mean falls in it.

Given 99% confidence interval : $4.1$

Then, the correct interpretation : We are 99% confident that the interval from 4.1 to 5.6 actually does contain the true value of mu.

8 0

3 years ago

Joaquin and Trisha are playing a game in which the lower median wins the game. Their scores are shown below. Joaquin's scores: 7

alekssr [168]

Answer:

Step-by-step explanation:

Given

Joaquin's score is $75,72,85,62,58,91$

and Trisha's score is $92,90,55,76,91,74$

Arranging score in order of value we get

Joaquin's : $58,62,72,75,85,91$

Trisha's : $55,74,76,90,91,92$

as no of values is even therefore their median is

Joaquin's $=\frac{72+75}{2}=73.5$

Trisha's $=\frac{76+90}{2}=83$

Therefore median of Joaquin's is lower

Thus Joaquin wins the game

5 0

3 years ago

Read 2 more answers

7 * ( -15 ):<br><br> por favor, es urgente):

Mademuasel [1]

Answer: From what I understand you want me to do 7*(-15)??? and the answer to that is -105.

Step-by-step explanation: All you have to do is 7*-15

8 0

3 years ago

If g(x)=4x^2-16 we’re shifted 9 united to the right and 1 down, what would the new equation be?

Inessa05 [86]

Shown in the graph

<h2>Explanation:</h2>

Using graph tools we can graph the function:

$g(x)=4x^2-16$

which is the red graph shown below. As you can see, this is a parabola. The rule for vertical and horizontal shifts is as follows:

$Let \ c \ be \ a \ positive \ real \ number. \ \mathbf{Vertical \ and \ horizontal \ shifts} \\ in \ the \ graph \ of \ y=f(x) \ are \ represented \ as \ follows:$

$\bullet \ Vertical \ shift \ c \ units \ \mathbf{upward}: \\ h(x)=f(x)+c \\ \\ \bullet \ Vertical \ shift \ c \ units \ \mathbf{downward}: \\ h(x)=f(x)-c \\ \\ \bullet \ Horizontal \ shift \ c \ units \ to \ the \ right \ \mathbf{right}: \\ h(x)=f(x-c) \\ \\ \bullet \ Horizontal \ shift \ c \ units \ to \ the \ left \ \mathbf{left}: \\ h(x)=f(x+c)$

Therefore, If we shift the red graph 9 units to the right and 1 down, our new function (let's call it $h(x)$ ) will be:

$h(x)=4\left(x-9\right)^{2}-16-1 \\ \\ Simplifying: \\ \\ h(x)=4\left(x-9\right)^{2}-17$

This graph is the blue graph below. Let's verify the transformation taking the vertex of the red graph:

$(0,-16)$

By translating the 9 units to the right and 1 down the vertex is also translated by the same rule, so:

$New \ vertex: \\ \\ (0+9,-16-1) \\ \\ (9,-17)$

<h2>Learn more:</h2>

Cubic function: brainly.com/question/13773618#

#LearnWithBrainly

5 0

3 years ago

A school administrator will assign each student in a group of n students to one of m classrooms. If 3 < m < 13 < n, is

tatiyna

Answer:

Hence to get same number of students in each classroom,the sufficient condition is that assign 13n students to each classroom.

Step-by-step explanation:

Given:

There are m classrooms and n be the students

3<m<13<n.

To Find:

Whether it is possible to assign each of n students to one of m classrooms with same no.of students.

Solution:

This problem is related to p/q form has to be integer in order to get same no of students assigned to the classroom.

As similar as ,n/m ratio

So 1st condition is that,

If it is possible to assign the n/m must be integer and n should be multiple of m,

when we assign 3n students to m classrooms ,we cannot say that 3n/m= integer so that n is greater than 13 i.e n=14 and m=6

hence they are not multiple of each other so they will not make same students in each classrooms.

Otherwise,n=14 and m=7 they will give same number but this condition is not sufficient condition to assign the student.

So 2nd condition is that ,

When we assign 13n students to m classrooms, as 13 is prime number and

3<m<13 which implies the 13n/m to be integer so n and m must be multiple of each other.

Suppose n=20 and m=5 classrooms

then 13*20=260 ,

260/5=52 students in each classroom,

4 0

3 years ago