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Kay [80]
3 years ago
7

True or False? The output values are the x-coordinates.

Mathematics
2 answers:
kap26 [50]3 years ago
6 0
False. it would be y.
Novay_Z [31]3 years ago
6 0

Answer:

The answer is false because if you have a set of ordered pairs, you can find the domain by listing all of the input values, which are the x-coordinates. And to find the range, list all of the output values, which are the y-coordinates. Hope this helps!!  :))

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The length of overline CD is 12 units. C^ prime D^ prime is the image of overline CD under a dilation with a scale factor of n.
Rufina [12.5K]

Answer:

A, D , and E

Step-by-step explanation:

We have that:

\overline{CD} = 12 \: units

\overline{C'D'}

is the image of CD after a dilation of scale factor n.

We use the relation between the image length and object length:

\overline{C'D'} = n \times \overline{CD}

Option A

If n=3/2, then

\overline{C'D'}  =  \frac{3}{2}  \times 12 = 3 \times 6 = 18 \: units

This is true.

Option B

If n=4, then

\overline{C'D'} = 4 \times 12 = 36 \: units

This is false.

Option C

If n=8, then

\overline{C'D'} = 8 \times 12 = 96 \: units

This too is false.

Option D

If n=2, then

\overline{C'D'} = 2 \times 12 = 24 \: units

This is true

Option E

If n=3/4, then

\overline{C'D'} =  \frac{3}{4}  \times  12

\overline{C'D'} =  3 \times 3 = 9 \: units

This is also true.

4 0
3 years ago
Find the value. please help!
ZanzabumX [31]
A= 2.5
b= 2
LM/ON= 15
LO/MN= 5
8 0
2 years ago
Write the equation x2+24x+60=0 in the form (x+p)2=q
navik [9.2K]

Answer:

(x+12)^2=84

Step-by-step explanation:

x^2+24x+60=0

x^2+24x=-60

x^2+24x+(24/2)^2=-60+(24/2)^2

(x+24/2)^2=84

(x+12)^2=84 (ans)

4 0
2 years ago
How do you simplify this –34–23a+16a
Dafna11 [192]

Answer:

-34 -7a

Step-by-step explanation:

–34–23a+16a

Combine like terms

-34 -7a

8 0
3 years ago
Read 2 more answers
Square of a standard normal: Warmup 1.0 point possible (graded, results hidden) What is the mean ????[????2] and variance ??????
LenaWriter [7]

Answer:

E[X^2]= \frac{2!}{2^1 1!}= 1

Var(X^2)= 3-(1)^2 =2

Step-by-step explanation:

For this case we can use the moment generating function for the normal model given by:

\phi(t) = E[e^{tX}]

And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:

\phi(t) = C \int_{R} e^{tx} e^{-\frac{x^2}{2}} dx = C \int_R e^{-\frac{x^2}{2} +tx} dx = e^{\frac{t^2}{2}} C \int_R e^{-\frac{(x-t)^2}{2}}dx

And we have that the moment generating function can be write like this:

\phi(t) = e^{\frac{t^2}{2}

And we can write this as an infinite series like this:

\phi(t)= 1 +(\frac{t^2}{2})+\frac{1}{2} (\frac{t^2}{2})^2 +....+\frac{1}{k!}(\frac{t^2}{2})^k+ ...

And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:

E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....]

E[e^{tX}]= 1+ E[X]t +\frac{1}{2}E[X^2]t^2 +....+\frac{1}{n1}E[X^n] t^n+...

and we can use the property that the convergent power series can be equal only if they are equal term by term and then we have:

\frac{1}{(2k)!} E[X^{2k}] t^{2k}=\frac{1}{k!} (\frac{t^2}{2})^k =\frac{1}{2^k k!} t^{2k}

And then we have this:

E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...

And then we can find the E[X^2]

E[X^2]= \frac{2!}{2^1 1!}= 1

And we can find the variance like this :

Var(X^2) = E[X^4]-[E(X^2)]^2

And first we find:

E[X^4]= \frac{4!}{2^2 2!}= 3

And then the variance is given by:

Var(X^2)= 3-(1)^2 =2

7 0
3 years ago
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