Ask question

Ask question

All categories

3 years ago

10

Please help me please !!

1 answer:

Artist 52 [7]3 years ago

5 0

The range is: 35-60
The IQR is: 11
This value is the first 25%

IQR= interquartile range

Hope this helps ;)

You might be interested in

Is anyone able to see if there is a counterexample for the following argument in logic?

Colt1911 [192]

if A=B

B=/=C

No A=C

All C are not A

6 0

3 years ago

The quotient of 5 plus a number down and negative 2 is 14

levacccp [35]

Answer:

quotient means division

(5 + d) / -2 = 14

Step-by-step explanation:

quotient means division

(5 + d) / -2 = 14

6 0

3 years ago

The quotient of a number and 6decreased by 10 is equal to 2??

Bas_tet [7]

The number equals -48

7 0

3 years ago

Which of te following statemnet are true

valkas [14]

Answer:

B

Step-by-step explanation:

Clearly BD is shorter than AE so that is false.

BD is clearly on the opposite end of EC so it is not bisecting, as bisecting would be DE bisecting AC

AC has to be perpendicular to something to be a perpendicular bisector however it is a straight line.

6 0

3 years ago

Read 2 more answers

The partial fraction decomposition of 9 x 11 6 x 2 23 x 21 can be written in the form of f ( x ) 2 x 3 g ( x ) 3 x 7 , where

svlad2 [7]

Answer:

$f(x) = -1$

$g(x) =6$

$\frac{-1}{2x + 3} + \frac{6}{3x + 7}$

Step-by-step explanation:

The question is unreadable, however the real polynomial is:

The polynomial fraction is:

$P = \frac{9x + 11}{6x^2 + 23x + 21}$

And the decomposition is:

$\frac{f(x)}{2x + 3} + \frac{g(x)}{3x + 7}$

The solution is as follows:

$P = \frac{f(x)}{2x + 3} + \frac{g(x)}{3x + 7}$

Substitute the expression for P

$\frac{9x + 11}{6x^2 + 23x + 21} = \frac{f(x)}{2x + 3} + \frac{g(x)}{3x + 7}$

Expand the numerator of the polynomial

$\frac{9x + 11}{(2x + 3)(3x + 7)} = \frac{f(x)}{2x + 3} + \frac{g(x)}{3x + 7}$

Take LCM

$\frac{9x + 11}{(2x + 3)(3x + 7)} = \frac{f(x)*(3x + 7) + g(x)*(2x + 3)}{(2x + 3)(x + 7)}$

Cancel out both denominators

$9x + 11} = f(x)*(3x + 7) + g(x)*(2x + 3)$

Represent f(x) as A and g(x) as B

$9x + 11} = A*(3x + 7) + B*(2x + 3)$

Open bracket

$9x + 11} = 3Ax + 7A + 2Bx + 3B$

$9x + 11} = 3Ax + 2Bx+ 7A + 3B$

$9x + 11} = (3A + 2B)x+ 7A + 3B$

By comparison:

$3A + 2B = 9$ ---- (1)

$7A + 3B =11$ ---- (2)

Make B the subject in (1)

$B = \frac{9 - 3A}{2}$

Substitute $B = \frac{9 - 3A}{2}$ in (2)

$7A + 3(\frac{9 - 3A}{2}) = 11$

Multiply through by 2

$2*7A +2* 3(\frac{9 - 3A}{2}) = 11*2$

$14A + 3(9 - 3A) = 22$

$14A + 27 - 9A = 22$

Collect Like Terms

$14A - 9A = 22-27$

$5A = -5$

$A = \frac{-5}{5}$

$A = -1$

Recall that:

$B = \frac{9 - 3A}{2}$

$B = \frac{9 - 3 * -1}{2}$

$B = \frac{9 + 3}{2}$

$B = \frac{12}{2}$

$B = 6$

A = -1 and B = 6

$3A + 2B = 9$ ---- (1)

$7A + 3B =11$ ---- (2)

So:

$f(x) = A$

$f(x) = -1$

$g(x) =B$

$g(x) =6$

And the decomposition is:

$\frac{-1}{2x + 3} + \frac{6}{3x + 7}$

6 0

2 years ago