Assuming that A and B represent any two sets, identify the statement as either always true or not always true.

Suppose we are testing the null hypothesis H 0 mu=20 and the alternative H iu =20 , normal population with sigma=6 A random samp

cupoosta [38]

Answer:

The value of the test statistic is $z = -1.5$

Step-by-step explanation:

The formula for the test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the statistic, $\mu$ is the mean, $\sigma$ is the standard deviation and n is the number of observations.

In this problem, we have that:

$\mu = 20, X = 17, \sigma = 6, n = 9$

So

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{17 - 20}{\frac{6}{\sqrt{3}}}$

$z = -1.5$

The value of the test statistic is $z = -1.5$

8 0

3 years ago

If 3x-2+5x=2,find the value of 3x

BlackZzzverrR [31]

Dont cheat do it your self

5 0

3 years ago

Determine the number of real solutions for each of the given equations. Equation Number of Solutions y = -3x2 + x + 12 y = 2x2 -

rosijanka [135]

Answer:

Step-by-step explanation:

Our equations are

$y = -3x^2 + x + 12\\y = 2x^2 - 6x + 5\\y = x^2 + 7x - 11\\y = -x^2 - 8x - 16\\$

Let us understand the term Discriminant of a quadratic equation and its properties

Discriminant is denoted by D and its formula is

$D=b^2-4ac\\$

Where

a= the coefficient of the $x^{2}$

b= the coefficient of $x$

c = constant term

Properties of D: If D

i) D=0 , One real root

ii) D>0 , Two real roots

iii) D<0 , no real root

Hence in the given quadratic equations , we will find the values of D Discriminant and evaluate our answer accordingly .

Let us start with

$y = -3x^2 + x + 12\\a=-3 , b =1 , c =12\\D=1^2-4*(-3)*(12)\\D=1+144\\D=145\\D>0 \\$

Hence we have two real roots for this equation.

$y = 2x^2 - 6x + 5\\$

$y = 2x^2 - 6x + 5\\a=2,b=-6,c=5\\D=(-6)^2-4*2*5\\D=36-40\\D=-4\\D$

Hence we do not have any real root for this quadratic

$y = x^2 + 7x - 11\\a=1,b=7,-11\\D=7^2-4*1*(-11)\\D=49+44\\D=93\\$

Hence D>0 and thus we have two real roots for this equation.

$y = -x^2 - 8x - 16\\a=-1,b=-8,c=-16\\D=(-8)^2-4*(-1)*(-16)\\D=64-64\\D=0\\$

Hence we have one real root to this quadratic equation.

7 0

2 years ago

Parker is laying pavers to build a patio in his backyard. Each square paver measures 2 feet on each side. He would like the pati

MatroZZZ [7]

Answer:

32 pavers

Step-by-step explanation:

step 1

Find out the area of one square paver

The area of a square is

$A=s^{2}$

where

s is the length side of the square

we have

$s=2\ ft$

substitute

$A=2^{2}\\A=4\ ft^{2}$

step 2

Find out the area of the rectangular patio

we know that

The area of a rectangle is

$A=LW$

we have

$L=16\ ft\\W=8\ ft$

substitute

$A=(16)(8)\\A=128\ ft^2$

step 3

Find out the number of pavers needed to build the patio

Divide the area of the rectangular patio by the area of one paver

$128\ ft^2/4\ ft^{2}=32\ pavers$

5 0

2 years ago

Find an equation of the circle that satisfies the given conditions.

hodyreva [135]

The equation of a circle:
$(x-h)^2+(y-k)^2=r^2$
(h,k) - the coordinates of the centre
r - the radius

The midpoint of the diameter is the centre of a circle.
The coordinates of the midpoint:
$(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$
(x₁,y₁), (x₂,y₂) - the coordinates of endpoints

$P(-1,1) \\
x_1=-1 \\ y_1=1 \\ \\ Q(5,9) \\ x_2=5 \\ y_2=9 \\ \\
\frac{x_1+x_2}{2}=\frac{-1+5}{2}=\frac{4}{2}=2 \\ \frac{y_1+y_2}{2}=\frac{1+9}{2}=\frac{10}{2}=5$

The centre of the circle is (2,5).

The radius is the distance between an endpoint of the diameter and the centre.
The formula for distance:
$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$(-1,1) \\ x_1=-1 \\ y_1=1 \\ \\ (2,5) \\ x_2=2 \\ y_2=5 \\ \\ d=\sqrt{(2-(-1))^2+(5-1)^2}=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$

The radius is 5.

$(x-2)^2+(y-5)^2=5^2 \\
\boxed{(x-2)^2+(y-5)^2=25}$

4 0

2 years ago

Assuming that A and B represent any two​ sets, identify the statement as either always true or not always true.

Assuming that A and B represent any two sets, identify the statement as either always true or not always true.