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3 years ago

Which ordered pair is a solution of the equation?

Mathematics

2 answers:

seraphim [82]3 years ago

8 0

Answer:

B.- Only (-3,5)

Step-by-step explanation:

In an ordered pair the first coordinate is the $x$ coordinate, and the second coordinate corresponds to the $y$ coordinate.

Then we must substitute the given value of $x$ in the equation and verify that we get in return the defined value for $y$ .

the options are

$(-2,4)$

in this one $x = -2$ , and $y=4$ , so let's confirm:

$y=-3x-4\\y=-3(-2)-4\\y=-6-4\\y=2$ this is not a solution since the $y$ values are not the same.

$(-3,5)$

in this one $x= -3$ and $y=5$ , let's also confirm:

$y=-3(-3)-4\\y=9-4\\y=5$ in this one the values for $y$ are equal, this is a solution for the equation.

frozen [14]3 years ago

3 0

It’s B because 5 = -3(-3) - 4
5=9-4
Just plug in the numbers and see what works

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Suppose that the proportions of blood phenotypes in a particular population are as follows: A B AB O 0.48 0.13 0.03 0.36 Assumin

Serggg [28]

Answer:

P(O and O) =0.1296

P=0.3778

Step-by-step explanation:

Given that

blood phenotypes in a particular population

A=0.48

B=0.13

AB=0.03

O=0.36

As we know that when A and B both are independent that

P(A and B)= P(A) X P(B)

The probability that both phenotypes O are in independent:

P(O and O)= P(O) X P(O)

P(O and O)= 0.36 X 0.36 =0.1296

P(O and O) =0.1296

The probability that the phenotypes of two randomly selected individuals match:

Here four case are possible

P=P(A and A)+P(B and B)+P(AB and AB)+P(O and O)

P=0.48 x 0.48 + 0.13 x 0.13 + 0.03 x 0.03 + 0.36 x 0.36

P=0.3778

7 0

3 years ago

Arrange the entries of matrix A in increasing order of their cofactors values

givi [52]

To find the cofactor of

$A=\left[\begin{array}{ccc}7&5&3\\-7&4&-1\\-8&2&1\end{array}\right]$

We cross out the Row and columns of the respective entries and find the determinant of the remaining $2\times 2$ matrix with the alternating signs.

$Ac_{11}=\left|\begin{array}{ccc}4&-1\\2&1\end{array}\right|$

$Ac_{11}=4\times 1- -1\times 2$

$Ac_{11}=4+ 2$

$Ac_{11}=6$

$Ac_{12}=-\left|\begin{array}{ccc}-7&-1\\-8&1\end{array}\right|$

$Ac_{12}=-(-7\times 1- -1\times -8)$

$Ac_{12}=-(-7- 8)$

$Ac_{12}=15$

$Ac_{21}=-\left|\begin{array}{ccc}5&3\\2&1\end{array}\right|$

$Ac_{21}=-(5\times 1- 3\times 2)$

$Ac_{21}=-(5-6)$

$Ac_{21}=1$

$A_c{23}=-\left|\begin{array}{ccc}7&5\\-8&2\end{array}\right|$

$Ac_{23}=-(7\times 2 -8\times 5)$

$Ac_{23}=-(14-40)$

$Ac_{23}=26$

$A_c{31}=\left|\begin{array}{ccc}5&3\\4&-1\end{array}\right|$

$Ac_{31}=5\times -1 -4\times 3$

$Ac_{31}=-5-12$

$Ac_{31}=-17$

$A_c{33}=\left|\begin{array}{ccc}7&5\\-7&4\end{array}\right|$

$Ac_{33}=7\times 4- -7\times 5$

$Ac_{33}=28+35$

$Ac_{33}=63$

Therefore in increasing order, we have;

$Ac_{31}=-17,Ac_{21}=1,Ac_{11}=6,Ac_{23}=26,Ac_{12}=15, Ac_{33}=63$

7 0

3 years ago

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Digiron [165]

Answer:

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Step-by-step explanation:

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Iteru [2.4K]

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Answer:

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