Answer: a11 = 3, a12 = 7, a21 = 1, a22 = -4
<u>Explanation:</u>
3x + 7y = 20
x - 4y = 9
A = ![\left[\begin{array}{ccc}3&7&20\\1&-4&9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%267%2620%5C%5C1%26-4%269%5Cend%7Barray%7D%5Cright%5D)
<u>column 1 </u> <u>column 2</u>
row 1: (a11) 3 (a12) 7
row 2: (a21) 1 (a22) -4
Hello There!
They're in different places the 3 in the ones place couldn't equal as much as the three in the thousands place. It all depends on where the numbers are in relation to the decimal.
Answer:
you want 4 correct and 16 incorrect
there are 20 questions
each question has four answers, so
P(right answer) = 1/4
P(wrong answer) = 3/4
----
Since you want 4 correct of 20 we have a combination of 20C4
This is a binomial problem where p = 1/4, q = 3/4 and we get
(20 "choose" 4)*(probability correct)^(number correct)*(probability incorrect)^(number incorrect)
putting numbers in we get
(20c4)*(1/4)^4*(3/4)^16
This gives us
~ .189685
Step-by-step explanation:
