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Xelga [282]
3 years ago
10

Beg someone help please

Mathematics
2 answers:
inysia [295]3 years ago
8 0
A=-9 B=-8 C=7 D=16

Explanation: just plug the numbers into the x spot
dem82 [27]3 years ago
6 0

Answer:

a=-26,b=-10,c=-1,d=-2

Step-by-step explanation:

when x=-3

y=-(-3)^2 +4(-3)-5=a

a=-9-12-5

=-14-12

=-26

when x=-1

y=-(-1)^2 +4(-1)-5=b

=-1-4-5

b=-10

when x=2

y=-(2)^2 +4(2)-5=c

=-4+8-5

=-9+8

c=-1

when x=3

y=-(3)^2 +4(3)-5=d

=-9+12-5

=-14+12

d=-2

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If the mean of this distribution is 13.7, which could NOT be the median?
Dmitriy789 [7]

Answer:

b

Step-by-step explanation:

5 0
3 years ago
Solve the system of linear equations using elimination.
N76 [4]

Answer:

C (-2,1)

Step-by-step explanation:

−2x − y = 3 (1)

−9x − y = 17 (2)

(1) y = -2x - 3

(2) y = -9x - 17

-2x - 3 = -9x - 17

7x = -14

x = -2

y = -2(-2) - 3

y = 4 - 3

y = 1

8 0
3 years ago
What is h(x) = 7 + 10x + x2 written in vertex form? h(x) = (x – 25)2 – 18 h(x) = (x – 5)2 + 32 h(x) = (x + 5)2 – 18 h(x) = (x +
muminat

Answer:c

Step-by-step explanation:

6 0
3 years ago
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
Which is the equivalent of 12.42 written in DMS form?
nika2105 [10]
12.42

12°+.42(60)

12°25.2'

12°25'+.2(60)

12°25'12"
3 0
3 years ago
Read 2 more answers
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