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3 years ago

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If the range of f (x) = StartRoot m x EndRoot and the range of g (x) = m StartRoot x EndRoot are the same, which statement is tr

ue about the value of m? m can only equal 1. m can be any positive real number. m can be any negative real number. m can be any real number.

2 answers:

andriy [413]3 years ago

7 0

Answer:

The answer is B on EDGE 2020

Alona [7]3 years ago

4 0

Answer:B

Step-by-step explanation:

You might be interested in

A survey is conducted to find out whether people in metropolitan areas obtain their news from television (Event T), an newspaper

MrMuchimi

Answer:

e) P ( R' | T ) = 0.62857

f) P ( T' | N ) = 0.32258

g) P ( R' | N ) = 0.70968

h) P ( T' | N&R ) = 5/6

Step-by-step explanation:

Given:

- The probability of Television as news source P ( T ) = 0.7

- The probability of Newspaper as news source P ( N ) = 0.62

- The probability of Radio as news source P ( R ) = 0.46

- The probability of Television & Newspaper as news source P (T&N) = 0.42

- The probability of Television & Radio as news source P (T&R) = 0.26

- The probability of Radio & Newspaper as news source P (R&N) = 0.18

- The probability of all 3 as news source P ( T & R & N ) = 0.03

Find:

Given that TV is a news source, what is the probability that radio is not a news source?

Given that newspaper is a news source, what is the probability that TV is not a news source?

Given that newspaper is a news source, what is the probability that radio is not a news source?

Given that both newspaper and radio are news sources, what is the probability that TV is not a news source?

Solution:

- We will first compute the individual probability of each event occurring alone.

P (Television is the only news source) = P(T) - P(T&R) - P(T&N) + P(T&N&R)

P ( Only Television ) = P ( only T ) = 0.7 - 0.42 - 0.26 + 0.03 = 0.05

P (Newspaper is the only news source) = P(N)-P(N&R)-P(T&N)+P(T&N&R)

P ( Only Newspaper ) = P ( only N ) = 0.62 - 0.18 - 0.42 + 0.03 = 0.05

P (Radio is the only news source) = P(R) - P(T&R) - P(N&R) + P(T&N&R)

P ( Only Radio ) = P ( only R ) = 0.46 - 0.26 - 0.18 + 0.03 = 0.05

- Now for part e)

We are asked for a conditional probability of the form as follows:

P ( R' | T ) = P ( R' & T ) / P ( T )

First compute the probability that next news source is not Radio provided it is already a source of TV.

P ( R' & T ) = P( only T ) + P ( only T & N ) = 0.05 + 0.42 - 0.03 = 0.44

Hence,

P ( R' | T ) = 0.44 / 0.7

P ( R' | T ) = 0.62857

- Now for part f)

We are asked for a conditional probability of the form as follows:

P ( T' | N ) = P ( T' & N ) / P ( N )

First compute the probability that next news source is not TV provided it is already a source of Newspaper.

P ( T' & N ) = P( only N ) + P ( only R & N ) = 0.05 + 0.18 - 0.03 = 0.2

Hence,

P ( T' | N ) = 0.2 / 0.62

P ( T' | N ) = 0.32258

- Now for part g)

We are asked for a conditional probability of the form as follows:

P ( R' | N ) = P ( R' & N ) / P ( N )

First compute the probability that next news source is not Radio provided it is already a source of Newspaper.

P ( R' & N ) = P( only N ) + P ( only T & N ) = 0.05 + 0.42 - 0.03 = 0.44

Hence,

P ( R' | N ) = 0.44 / 0.62

P ( R' | N ) = 0.70968

- Now for part h)

We are asked for a conditional probability of the form as follows:

P ( T' | N&R ) = P ( T' & N & R) / P ( N & R )

First compute the probability that next news source is not TV provided it is already a source of both radio and Newspaper.

P ( T' & N & R) = P ( only N & R ) = 0.18 - 0.03 = 0.15

Hence,

P ( T' | N&R ) = 0.15 / 0.18

P ( T' | N&R ) = 5/6

6 0

3 years ago

Triangle A E C is shown. Line segment B D is drawn near point C to form triangle B D C.

AVprozaik [17]

∠BDC and ∠AED are right angles, is a piece of additional information is appropriate to prove △ CEA ~ △ CDB

Triangle AEC is shown. Line segment B, D is drawn near point C to form triangle BDC.

<h3> What are Similar triangles?</h3>

Similar triangles, are those triangles which have similar properties,i.e. angles and proportionality of sides.

Image is attached below,
as shown in figure
∡ACE = ∡BCD ( common angle )
∡AED = ∡BDC ( since AE and BD are perpendicular to same line EC and make right angles as E and C)
∡EAC =- ∡DBC ( corresponding angles because AE and BD are parallel lines)

Thus, △CEA ~ △CDB , because of the two perpendiculars AE and BD.

Learn more about similar triangles here:
brainly.com/question/25882965

#SPJ1

3 0

1 year ago

Question 8 (06.02 LC)

kodGreya [7K]

Question 8.
The best answer would be A, two times y divided by 7.

Question 9.
The distributive property. Or A.

Question 10.
5(2+y) would be the answer. So, D.

Hope this helps!

8 0

2 years ago

PLS I NEED HELP WITH MY EXAM

Neporo4naja [7]

Answer:

F. a1 = –5, an = 5an-1

Step-by-step explanation:

a1 = -5 <---- Answer

r = -25/-5 = 5

an = r(an - 1)

an = 5(an - 1) <---- Answer

Hope this helps!

5 0

3 years ago

Read 2 more answers

The plane x + y + z = 12 intersects paraboloid z = x^2 + y^2 in an ellipse.(a) Find the highest and the lowest points on the ell

emmasim [6.3K]

Answer:

a)

Highest (-3,-3)

Lowest (2,2)

b)

Farthest (-3,-3)

Closest (2,2)

Step-by-step explanation:

To solve this problem we will be using Lagrange multipliers.

a)

Let us find out first the restriction, which is the projection of the intersection on the XY-plane.

From x+y+z=12 we get z=12-x-y and replace this in the equation of the paraboloid:

$\bf 12-x-y=x^2+y^2\Rightarrow x^2+y^2+x+y=12$

completing the squares:

$\bf x^2+y^2+x+y=12\Rightarrow (x+1/2)^2-1/4+(y+1/2)^2-1/4=12\Rightarrow\\\\\Rightarrow (x+1/2)^2+(y+1/2)^2=12+1/2\Rightarrow (x+1/2)^2+(y+1/2)^2=25/2$

and we want the maximum and minimum of the paraboloid when (x,y) varies on the circumference we just found. That is, we want the maximum and minimum of

$\bf f(x,y)=x^2+y^2$

subject to the constraint

$\bf g(x,y)=(x+1/2)^2+(y+1/2)^2-25/2=0$

Now we have

$\bf \nabla f=(\displaystyle\frac{\partial f}{\partial x},\displaystyle\frac{\partial f}{\partial y})=(2x,2y)\\\\\nabla g=(\displaystyle\frac{\partial g}{\partial x},\displaystyle\frac{\partial g}{\partial y})=(2x+1,2y+1)$

Let $\bf \lambda$ be the Lagrange multiplier.

The maximum and minimum must occur at points where

$\bf \nabla f=\lambda\nabla g$

that is,

$\bf (2x,2y)=\lambda(2x+1,2y+1)\Rightarrow 2x=\lambda (2x+1)\;,2y=\lambda (2y+1)$

we can assume (x,y)≠ (-1/2, -1/2) since that point is not in the restriction, so

$\bf \lambda=\displaystyle\frac{2x}{(2x+1)} \;,\lambda=\displaystyle\frac{2y}{(2y+1)}\Rightarrow \displaystyle\frac{2x}{(2x+1)}=\displaystyle\frac{2y}{(2y+1)}\Rightarrow\\\\\Rightarrow 2x(2y+1)=2y(2x+1)\Rightarrow 4xy+2x=4xy+2y\Rightarrow\\\\\Rightarrow x=y$

Replacing in the constraint

$\bf (x+1/2)^2+(x+1/2)^2-25/2=0\Rightarrow (x+1/2)^2=25/4\Rightarrow\\\\\Rightarrow |x+1/2|=5/2$

from this we get

<em>x=-1/2 + 5/2 = 2 or x = -1/2 - 5/2 = -3 </em>

<em> </em>

and the candidates for maximum and minimum are (2,2) and (-3,-3).

Replacing these values in f, we see that

f(-3,-3) = 9+9 = 18 is the maximum and

f(2,2) = 4+4 = 8 is the minimum

b)

Since the square of the distance from any given point (x,y) on the paraboloid to (0,0) is f(x,y) itself, the maximum and minimum of the distance are reached at the points we just found.

We have then,

(-3,-3) is the farthest from the origin

(2,2) is the closest to the origin.

3 0

3 years ago