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3 years ago

8

Can anyone help me out please? A rectangle has an area of (x^3- 2x^2 +3x- 18) square meters and a width of (x-3) meters. Find it

’s length, in terms of X.

1 answer:

adelina 88 [10]3 years ago

8 0

I have solved the answer in the pic below, hope it helpss!!!!

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Solve the following system of equations express your answers as an ordered pair in the format (a,b) with no spaces between the n

EleoNora [17]

Double the first equation: 4x+14y=-2 and subtract the second: 17y=17, so y=1. 2x=-7y-1=-7-1=-8, so x=-8/2=-4.
The answer is x=-4 and y=1.

4 0

3 years ago

Read 2 more answers

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

Factor completely: 2x^3 – 32x

anastassius [24]

$2x^3-32x= \\
2x(x^2-16)= \\
2x(x^2-4^2)= \\
\boxed{2x(x+4)(x-4)} \Leftarrow
\hbox{answer B}$

7 0

4 years ago

The weights of 67 randomly selected axles were found to have a variance of 3.85. Construct the 80% confidence interval for the p

VLD [36.1K]

Answer:

$3.13$

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 67

Variance = 3.85

We have to find 80% confidence interval for the population variance of the weights.

Degree of freedom = 67 - 1 = 66

Level of significance = 0.2

Chi square critical value for lower tail =

$\chi^2_{1-\frac{\alpha}{2}}= 51.770$

Chi square critical value for upper tail =

$\chi^2_{\frac{\alpha}{2}}= 81.085$

80% confidence interval:

$\dfrac{(n-1)S^2}{\chi^2_{\frac{\alpha}{2}}} < \sigma^2 < \dfrac{(n-1)S^2}{\chi^2_{1-\frac{\alpha}{2}}}$

Putting values, we get,

$=\dfrac{(67-1)3.85}{81.085} < \sigma^2 < \dfrac{(67-1)3.85}{51.770}\\\\=3.13$

Thus, (3.13,4.91) is the required 80% confidence interval for the population variance of the weights.

8 0

3 years ago

There are 30 computers in the media lab there are 4 activity booklets for every 2 computers in the media lab.

AURORKA [14]

Answer:

There are 60 activity booklets in the media lab.

Step-by-step explanation:

- 30/2 = 15

Why divide those numbers?

For every 2 computers in a lab with 30 computers means you have to divide

- 15x4 = 60

Why multiply those numbers?

4 activity booklets for every 2 computers in a lab of 30 computers, so multiply

I hope this helps.

7 0

4 years ago