Answer:
Option b is correct (8,13).
Step-by-step explanation:
7x - 4y = 4
10x - 6y =2
it can be represented in matrix form as![\left[\begin{array}{cc}7&-4\\10&-6\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{c}4\\2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D7%26-4%5C%5C10%26-6%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%5C%5C2%5Cend%7Barray%7D%5Cright%5D)
A=
X= ![\left[\begin{array}{c}x\\y\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D)
B= ![\left[\begin{array}{c}4\\2\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%5C%5C2%5Cend%7Barray%7D%5Cright%5D)
i.e, AX=B
or X= A⁻¹ B
A⁻¹ = 1/|A| * Adj A
determinant of A = |A|= (7*-6) - (-4*10)
= (-42)-(-40)
= (-42) + 40 = -2
so, |A| = -2
Adj A=
A⁻¹ =
/ -2
A⁻¹ = ![\left[\begin{array}{cc}3&-2\\5&-7/2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%26-2%5C%5C5%26-7%2F2%5Cend%7Barray%7D%5Cright%5D%20)
X= A⁻¹ B
X= ![\left[\begin{array}{cc}3&-2\\5&-7/2\end{array}\right] *\left[\begin{array}{c}4\\2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%26-2%5C%5C5%26-7%2F2%5Cend%7Barray%7D%5Cright%5D%20%2A%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%5C%5C2%5Cend%7Barray%7D%5Cright%5D)
X= ![\left[\begin{array}{c}(3*4) + (-2*2)\\(5*4) + (-7/2*2)\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D%283%2A4%29%20%2B%20%28-2%2A2%29%5C%5C%285%2A4%29%20%2B%20%28-7%2F2%2A2%29%5Cend%7Barray%7D%5Cright%5D)
X= ![\left[\begin{array}{c}12-4\\20-7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D12-4%5C%5C20-7%5Cend%7Barray%7D%5Cright%5D)
X= ![\left[\begin{array}{c}8\\13\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D8%5C%5C13%5Cend%7Barray%7D%5Cright%5D)
x= 8, y= 13
solution set= (8,13).
Option b is correct.
Answer:
Fraction 1/6. Decimal .6 or 1.6
Step-by-step explanation:
10 would be 50% so 1 is 5% 5x8 =40
Her experimental probability is 40%
Complete question :
The average daily volume of a computer stock in 2011 was p = 35.1 million shares, according to a reliable source. A stock analyst believes that the stock volume in 2014 is different from the 2011 level. Based on a random sample of 40 trading days in 2014, he finds the sample mean to be 30.9 million shares, with a standard deviation of s = 11.8 million shares. Test the hypotheses by constructing a 95% confidence interval. Complete parts (a) through (c) below. State the hypotheses for the test. Construct a 95% confidence interval about the sample mean of stocks traded in 2014.
Answer:
H0 : μ = 35.1 ;
H1 : μ < 35.1 ;
(26.488 ; 35.312)
Step-by-step explanation:
The hypothesis :
H0 : μ = 35.1
H1 : μ < 35.1
The confidence interval :
Xbar ± Margin of error
Xbar = 30.9
Margin of Error = Zcritical * s/sqrt(n)
Zcritical at 95% = 1.96
Margin of Error = 1.96 * (11.8/sqrt(40))
Margin of Error = 4.412
Lower boundary :
30.9 - 4.412 = 26.488
Upper boundary :
30.9 + 4.412 = 35.312
Confidence interval = (26.488 ; 35.312)
Since the population mean value exists within the interval, the we fail to reject the Null.