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belka [17]
3 years ago
14

Pls pls help!! there's 3 images. I WILL MARK YOU BRAINLIEST IF U GET IT RIGHT

Mathematics
1 answer:
Damm [24]3 years ago
6 0

Answer:

im not sure about the 1st one but sec one is 162 i think and im  i dont know the last eaither srry  

Step-by-step explanation:

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If the equation y² - (K-<br>2y + 2k +1 = 0<br>with equal roots find the value of k​
777dan777 [17]

Answer:

y² - (K-  2)y + 2k +1 = 0

equal roots means D=0

D= b^2 - 4ac

a=1, b= (k-2), c= 2k+1

so,

(k-2)^2 - 4(1)(2k+1) = 0

=> k^2 +4 - 8k -4 = 0

=> k^2 -8k = 0

=> k^2 = 8k

=> k= 8k/k

=> k = 8

Therefore the answer is k= 8

Hope it helps........

8 0
3 years ago
Last year your school had 1200 students enrolled, this year your school has135% of that amount enrolled. how many students are e
iren2701 [21]
1200 = 100%
35% of 1200 = 420
1200 + 420 = 1620

There are 1620 students enrolled this year.
7 0
2 years ago
Suppose I claim that the average monthly income of all students at college is at least $2000. Express H0 and H1 using mathematic
pishuonlain [190]

Answer:

For this case we want to test if the the average monthly income of all students at college is at least $2000. Since the alternative hypothesis can't have an equal sign thne the correct system of hypothesis for this case are:

Null hypothesis (H0): \mu \geq 2000

Alternative hypothesis (H1): \mu

And in order to test this hypothesis we can use a one sample t or z test in order to verify if the true mean is at least 200 or no

Step-by-step explanation:

For this case we want to test if the the average monthly income of all students at college is at least $2000. Since the alternative hypothesis can't have an equal sign thne the correct system of hypothesis for this case are:

Null hypothesis (H0): \mu \geq 2000

Alternative hypothesis (H1): \mu

And in order to test this hypothesis we can use a one sample t or z test in order to verify if the true mean is at least 2000 or no

7 0
2 years ago
Which is part interphase
Soloha48 [4]

Answer:

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
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
Read 2 more answers
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