You put x-5 equal to 0 and then add 5 to the other side
Work:
X-5=0
X=5
Answer:
(a)![x_1=-2,x_2=1](https://tex.z-dn.net/?f=x_1%3D-2%2Cx_2%3D1)
(b)![x_1=\frac{3}{4} ,x_2=-\frac{1}{2} ,x_3=\frac{1}{4}](https://tex.z-dn.net/?f=x_1%3D%5Cfrac%7B3%7D%7B4%7D%20%2Cx_2%3D-%5Cfrac%7B1%7D%7B2%7D%20%2Cx_3%3D%5Cfrac%7B1%7D%7B4%7D)
Step-by-step explanation:
(a) For using Cramer's rule you need to find matrix
and the matrix
for each variable. The matrix
is formed with the coefficients of the variables in the system. The first step is to accommodate the equations, one under the other, to get
more easily.
![2x_1+5x_2=1\\x_1+4x_2=2](https://tex.z-dn.net/?f=2x_1%2B5x_2%3D1%5C%5Cx_1%2B4x_2%3D2)
![\therefore A=\left[\begin{array}{cc}2&5\\1&4\end{array}\right]](https://tex.z-dn.net/?f=%5Ctherefore%20A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%265%5C%5C1%264%5Cend%7Barray%7D%5Cright%5D)
To get
, replace in the matrix A the 1st column with the results of the equations:
![B_1=\left[\begin{array}{cc}1&5\\2&4\end{array}\right]](https://tex.z-dn.net/?f=B_1%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%265%5C%5C2%264%5Cend%7Barray%7D%5Cright%5D)
To get
, replace in the matrix A the 2nd column with the results of the equations:
![B_2=\left[\begin{array}{cc}2&1\\1&2\end{array}\right]](https://tex.z-dn.net/?f=B_2%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%261%5C%5C1%262%5Cend%7Barray%7D%5Cright%5D)
Apply the rule to solve
:
![x_1=\frac{det\left(\begin{array}{cc}1&5\\2&4\end{array}\right)}{det\left(\begin{array}{cc}2&5\\1&4\end{array}\right)} =\frac{(1)(4)-(2)(5)}{(2)(4)-(1)(5)} =\frac{4-10}{8-5}=\frac{-6}{3}=-2\\x_1=-2](https://tex.z-dn.net/?f=x_1%3D%5Cfrac%7Bdet%5Cleft%28%5Cbegin%7Barray%7D%7Bcc%7D1%265%5C%5C2%264%5Cend%7Barray%7D%5Cright%29%7D%7Bdet%5Cleft%28%5Cbegin%7Barray%7D%7Bcc%7D2%265%5C%5C1%264%5Cend%7Barray%7D%5Cright%29%7D%20%3D%5Cfrac%7B%281%29%284%29-%282%29%285%29%7D%7B%282%29%284%29-%281%29%285%29%7D%20%3D%5Cfrac%7B4-10%7D%7B8-5%7D%3D%5Cfrac%7B-6%7D%7B3%7D%3D-2%5C%5Cx_1%3D-2)
In the case of B2, the determinant is going to be zero. Instead of using the rule, substitute the values of the variable
in one of the equations and solve for
:
![2x_1+5x_2=1\\2(-2)+5x_2=1\\-4+5x_2=1\\5x_2=1+4\\ 5x_2=5\\x_2=1](https://tex.z-dn.net/?f=2x_1%2B5x_2%3D1%5C%5C2%28-2%29%2B5x_2%3D1%5C%5C-4%2B5x_2%3D1%5C%5C5x_2%3D1%2B4%5C%5C%205x_2%3D5%5C%5Cx_2%3D1)
(b) In this system, follow the same steps,ust remember
is formed by replacing the 3rd column of A with the results of the equations:
![2x_1+x_2 =1\\x_1+2x_2+x_3=0\\x_2+2x_3=0](https://tex.z-dn.net/?f=2x_1%2Bx_2%20%3D1%5C%5Cx_1%2B2x_2%2Bx_3%3D0%5C%5Cx_2%2B2x_3%3D0)
![\therefore A=\left[\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right]](https://tex.z-dn.net/?f=%5Ctherefore%20A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%261%260%5C%5C1%262%261%5C%5C0%261%262%5Cend%7Barray%7D%5Cright%5D)
![B_1=\left[\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right]](https://tex.z-dn.net/?f=B_1%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%260%5C%5C0%262%261%5C%5C0%261%262%5Cend%7Barray%7D%5Cright%5D)
![B_2=\left[\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right]](https://tex.z-dn.net/?f=B_2%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%261%260%5C%5C1%260%261%5C%5C0%260%262%5Cend%7Barray%7D%5Cright%5D)
![B_3=\left[\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right]](https://tex.z-dn.net/?f=B_3%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%261%261%5C%5C1%262%260%5C%5C0%261%260%5Cend%7Barray%7D%5Cright%5D)
![x_1=\frac{det\left(\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{1(2)(2)+(0)(1)(0)+(0)(1)(1)-(1)(1)(1)-(0)(1)(2)-(0)(2)(0)}{(2)(2)(2)+(1)(1)(0)+(0)(1)(1)-(2)(1)(1)-(1)(1)(2)-(0)(2)(0)}\\ x_1=\frac{4+0+0-1-0-0}{8+0+0-2-2-0} =\frac{3}{4} \\x_1=\frac{3}{4}](https://tex.z-dn.net/?f=x_1%3D%5Cfrac%7Bdet%5Cleft%28%5Cbegin%7Barray%7D%7Bccc%7D1%261%260%5C%5C0%262%261%5C%5C0%261%262%5Cend%7Barray%7D%5Cright%29%7D%7Bdet%5Cleft%28%5Cbegin%7Barray%7D%7Bccc%7D2%261%260%5C%5C1%262%261%5C%5C0%261%262%5Cend%7Barray%7D%5Cright%29%7D%20%3D%5Cfrac%7B1%282%29%282%29%2B%280%29%281%29%280%29%2B%280%29%281%29%281%29-%281%29%281%29%281%29-%280%29%281%29%282%29-%280%29%282%29%280%29%7D%7B%282%29%282%29%282%29%2B%281%29%281%29%280%29%2B%280%29%281%29%281%29-%282%29%281%29%281%29-%281%29%281%29%282%29-%280%29%282%29%280%29%7D%5C%5C%20x_1%3D%5Cfrac%7B4%2B0%2B0-1-0-0%7D%7B8%2B0%2B0-2-2-0%7D%20%3D%5Cfrac%7B3%7D%7B4%7D%20%5C%5Cx_1%3D%5Cfrac%7B3%7D%7B4%7D)
![x_2=\frac{det\left(\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{(2)(0)(2)+(1)(0)(0)+(0)(1)(1)-(2)(0)(1)-(1)(1)(2)-(0)(0)(0)}{4} \\x_2=\frac{0+0+0-0-2-0}{4}=\frac{-2}{4}=-\frac{1}{2}\\x_2=-\frac{1}{2}](https://tex.z-dn.net/?f=x_2%3D%5Cfrac%7Bdet%5Cleft%28%5Cbegin%7Barray%7D%7Bccc%7D2%261%260%5C%5C1%260%261%5C%5C0%260%262%5Cend%7Barray%7D%5Cright%29%7D%7Bdet%5Cleft%28%5Cbegin%7Barray%7D%7Bccc%7D2%261%260%5C%5C1%262%261%5C%5C0%261%262%5Cend%7Barray%7D%5Cright%29%7D%20%3D%5Cfrac%7B%282%29%280%29%282%29%2B%281%29%280%29%280%29%2B%280%29%281%29%281%29-%282%29%280%29%281%29-%281%29%281%29%282%29-%280%29%280%29%280%29%7D%7B4%7D%20%5C%5Cx_2%3D%5Cfrac%7B0%2B0%2B0-0-2-0%7D%7B4%7D%3D%5Cfrac%7B-2%7D%7B4%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5C%5Cx_2%3D-%5Cfrac%7B1%7D%7B2%7D)
![x_3=\frac{det\left(\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)}=\frac{(2)(2)(0)+(1)(1)(1)+(0)(1)(0)-(2)(1)(0)-(1)(1)(0)-(0)(2)(1)}{4} \\x_3=\frac{0+1+0-0-0-0}{4}=\frac{1}{4}\\x_3=\frac{1}{4}](https://tex.z-dn.net/?f=x_3%3D%5Cfrac%7Bdet%5Cleft%28%5Cbegin%7Barray%7D%7Bccc%7D2%261%261%5C%5C1%262%260%5C%5C0%261%260%5Cend%7Barray%7D%5Cright%29%7D%7Bdet%5Cleft%28%5Cbegin%7Barray%7D%7Bccc%7D2%261%260%5C%5C1%262%261%5C%5C0%261%262%5Cend%7Barray%7D%5Cright%29%7D%3D%5Cfrac%7B%282%29%282%29%280%29%2B%281%29%281%29%281%29%2B%280%29%281%29%280%29-%282%29%281%29%280%29-%281%29%281%29%280%29-%280%29%282%29%281%29%7D%7B4%7D%20%5C%5Cx_3%3D%5Cfrac%7B0%2B1%2B0-0-0-0%7D%7B4%7D%3D%5Cfrac%7B1%7D%7B4%7D%5C%5Cx_3%3D%5Cfrac%7B1%7D%7B4%7D)
Answer:
The probability that 13 of them were very confident their major would lead to a good job is 1.08%.
Step-by-step explanation:
Given : A 2017 poll found that 56% of college students were very confident that their major will lead to a good job. If 15 college students are chosen at random.
To find : What's the probability that 13 of them were very confident their major would lead to a good job?
Solution :
Applying Binomial distribution,
![P(x)=^nC_x p^x q^{n-x}](https://tex.z-dn.net/?f=P%28x%29%3D%5EnC_x%20p%5Ex%20q%5E%7Bn-x%7D)
Here, p is the success p=56%=0.56
q is the failure ![q= 1-p=1-0.56=0.44](https://tex.z-dn.net/?f=q%3D%201-p%3D1-0.56%3D0.44)
n is the number of selection n=15
The probability that 13 of them were very confident their major would lead to a good job i.e. x=13
Substitute the values,
![P(13)=^{15}C_{13} (0.56)^{13} (0.44)^{15-13}](https://tex.z-dn.net/?f=P%2813%29%3D%5E%7B15%7DC_%7B13%7D%20%280.56%29%5E%7B13%7D%20%280.44%29%5E%7B15-13%7D)
![P(13)=\frac{15!}{13!2!}\times (0.56)^{13}\times (0.44)^{2}](https://tex.z-dn.net/?f=P%2813%29%3D%5Cfrac%7B15%21%7D%7B13%212%21%7D%5Ctimes%20%280.56%29%5E%7B13%7D%5Ctimes%20%280.44%29%5E%7B2%7D)
![P(13)=\frac{15\times 14}{2\times 1}\times (0.56)^{13}\times (0.44)^{2}](https://tex.z-dn.net/?f=P%2813%29%3D%5Cfrac%7B15%5Ctimes%2014%7D%7B2%5Ctimes%201%7D%5Ctimes%20%280.56%29%5E%7B13%7D%5Ctimes%20%280.44%29%5E%7B2%7D)
![P(13)=105\times (0.56)^{13}\times (0.44)^{2}](https://tex.z-dn.net/?f=P%2813%29%3D105%5Ctimes%20%280.56%29%5E%7B13%7D%5Ctimes%20%280.44%29%5E%7B2%7D)
![P(13)=0.0108](https://tex.z-dn.net/?f=P%2813%29%3D0.0108)
The probability that 13 of them were very confident their major would lead to a good job is 1.08%.
To solve the problem, we need to use the relation between the angles to find a relation involving x. Then we need to solve for x and use that value to determine the measure of the desired angle.
The diagram shows you ...
... ∠KIJ + ∠JIH = ∠KIH
The problem statement tells you ∠KIJ = ∠JIH, so we can substitute ∠JIH in the above equation to get
... ∠JIH + ∠JIH = ∠KIH
... (4x+3) + (4x+3) = (12x)
... 8x +6 = 12x
... 6 = 4x . . . . . . . subtract 8x
Since we want to know the value of 12x, we can go directly there
... 18 = 12x . . . . . . multiply the above equation by 3
m∠KIH = 18°
For the first choice x<2. This is seen by the open circle.
For the second, x ≤ -4.25. This is seen by the closed circle
For the third, the answer would be the first option.