Interquantile range x^2-5x+6=0

I need help graphing

steposvetlana [31]

Lmk if this helped if not im sorry

5 0

3 years ago

Any 10th grader solve it <br>for 50 points

kkurt [141]

Answer:

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0$ is proved for the sum of pth, qth and rth terms of an arithmetic progression are a, b,and c respectively.

Step-by-step explanation:

Given that the sum of pth, qth and rth terms of an arithmetic progression are a, b and c respectively.

First term of given arithmetic progression is A

and common difference is D

ie., $a_{1}=A$ and common difference=D

The nth term can be written as

$a_{n}=A+(n-1)D$

pth term of given arithmetic progression is a

$a_{p}=A+(p-1)D=a$

qth term of given arithmetic progression is b

$a_{q}=A+(q-1)D=b$ and

rth term of given arithmetic progression is c

$a_{r}=A+(r-1)D=c$

We have to prove that

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)=0$

Now to prove LHS=RHS

Now take LHS

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)$

$=\frac{A+(p-1)D}{p}\times (q-r)+\frac{A+(q-1)D}{q}\times (r-p)+\frac{A+(r-1)D}{r}\times (p-q)$

$=\frac{A+pD-D}{p}\times (q-r)+\frac{A+qD-D}{q}\times (r-p)+\frac{A+rD-D}{r}\times (p-q)$

$=\frac{Aq+pqD-Dq-Ar-prD+rD}{p}+\frac{Ar+rqD-Dr-Ap-pqD+pD}{q}+\frac{Ap+prD-Dp-Aq-qrD+qD}{r}$

$=\frac{[Aq+pqD-Dq-Ar-prD+rD]\times qr+[Ar+rqD-Dr-Ap-pqD+pD]\times pr+[Ap+prD-Dp-Aq-qrD+qD]\times pq}{pqr}$

$=\frac{Arq^{2}+pq^{2} rD-Dq^{2} r-Aqr^{2}-pqr^{2} D+qr^{2} D+Apr^{2}+pr^{2} qD-pDr^{2} -Ap^{2}r-p^{2} rqD+p^{2} rD+Ap^{2} q+p^{2} qrD-Dp^{2} q-Aq^{2} p-q^{2} prD+q^{2}pD}{pqr}$

$=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2}-pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}$

$=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2} -pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}$

$\neq 0$

ie., $RHS\neq 0$

Therefore $LHS\neq RHS$

ie., $\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0$

Hence proved

5 0

3 years ago

I need some help with this question :/

Masteriza [31]

You have one of the answers correct. The fraction 4/8 is equal to 1/2

Other answers are: 5/10, 10/20, and 6/12

If you reduced each of those fractions, you'd get to 1/2.

You can use a calculator to find that
4/8 = 0.5
1/2 = 0.5
5/10 = 0.5
10/20 = 0.5
6/12 = 0.5
All fractions mentioned lead to the same decimal value, so this confirms all of the fractions are equivalent to one another.

7 0

3 years ago

Read 2 more answers

Determine which function has the greatest rate of change as x approaches infinity.

Mekhanik [1.2K]

Answer: h(x) = 3*x^2 - 7*x + 8

Step-by-step explanation:

The rate of change of a function is equal to the derivate:

remember that a derivate of the form:

k(x) = a*x^n is k'(x) = n*a*x^(n-1)

Then we have:

f(x) = 2*x - 10

f'(x) = 1*2* = 2

g(x) = 16*x - 4

g'(x) = 1*16 = 16

h(x) = 3*x^2 - 7*x + 8

h'(x) = 2*3*x - 1*7 = 6*x - 7

So the only that increases as x increases is h(x), this means that the greates rate of change as x approaches inffinity is the rate of change of h(x)

3 0

3 years ago

Suppose a sample of 256 was taken from a population with a standard deviation of 64 centimeters. What is the margin of error for

Ganezh [65]

I think the answer is b

8 0

3 years ago

Read 2 more answers