Part A:
The probability that the birthday of three strangers were on Wednesday is given by
Part B:
The probability that the birthday of three strangers were on different days of the week is given by
Part C:
The probability that none of the three strangers were born on Saturday is given by
Brad's profit was 30%
Further explanation:
Profit is the money someone earns on selling a product on more than its actual cost.
The formula for profit is:
Given
Purchase price of energy drinks = £9.25
Selling price of one bottle=£1
Selling price of 12 bottles=1*12=£12
Profit= Selling Price - Purchse price
=12-9.25
=£2.75
Brad's profit was £2.75
Now,
Profit percentage= (Profit/Purchase Price)*100
=(2.75/9.25)*100
=0.30*100
=30%
Brad's profit was 30%
Keywords: Profit, loss
Learn more about profit and loss at:
#LearnwithBrainly
The difference between the city's highest and lowest recorded temperature
=47 - ( -13)
=60 °C
By implicit differentiation:
<span>(x(dy/dx) + y)e^(xy) = 0 </span>
<span>Note that when differentiating e^(xy), apply chain rule. When differentiating xy, use product rule. Also: When differentiating y w/respect to x, think of that as if you are differentiating f(x). </span>
<span>Then, substitute (1,ln(2)) and solve for dy/dx. </span>
<span>(1(dy/dx) + ln(2))e^(1ln(2)) = 0 </span>
<span>((dy/dx) + ln(2))e^(ln(2)) = 0 </span>
<span>Note that e^(ln(2)) = 2 since e and ln are inverse of each other. </span>
<span>2((dy/dx) + ln(2)) = 0 </span>
<span>dy/dx + ln(2) = 0 . . . . You get this expression by dividing both sides by 2 </span>
<span>dy/dx = -ln(2) . . . . . . .Subtract both sides by ln(2) </span>
<span>Therefore, dy/dx = -ln(2) </span>
<span>I hope this helps!</span>
The question is missing parts. Here is the complete question.
Let M = ![\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C-1%26-4%5Cend%7Barray%7D%5Cright%5D)
. Find 
and 
such that 
, where 
is the identity 2x2 matrix and 0 is the zero matrix of appropriate dimension.
Answer: 
Step-by-step explanation: Identity matrix is a sqaure matrix that has 1's along the main diagonal and 0 everywhere else. So, a 2x2 identity matrix is:
![\left[\begin{array}{cc}1&0\\0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D)
![M^{2} = \left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]](https://tex.z-dn.net/?f=M%5E%7B2%7D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C-1%26-4%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C-1%26-4%5Cend%7Barray%7D%5Cright%5D)
![M^{2}=\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]](https://tex.z-dn.net/?f=M%5E%7B2%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D31%2610%5C%5C-2%2615%5Cend%7Barray%7D%5Cright%5D)
Solving equation:
![\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]+c_{1}\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right] +c_{2}\left[\begin{array}{cc}1&0\\0&1\end{array}\right] =\left[\begin{array}{cc}0&0\\0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D31%2610%5C%5C-2%2615%5Cend%7Barray%7D%5Cright%5D%2Bc_%7B1%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C-1%26-4%5Cend%7Barray%7D%5Cright%5D%20%2Bc_%7B2%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0%260%5C%5C0%260%5Cend%7Barray%7D%5Cright%5D)
Multiplying a matrix and a scalar results in all the terms of the matrix multiplied by the scalar. You can only add matrices of the same dimensions.
So, the equation is:
![\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]+\left[\begin{array}{cc}6c_{1}&5c_{1}\\-1c_{1}&-4c_{1}\end{array}\right] +\left[\begin{array}{cc}c_{2}&0\\0&c_{2}\end{array}\right] =\left[\begin{array}{cc}0&0\\0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D31%2610%5C%5C-2%2615%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6c_%7B1%7D%265c_%7B1%7D%5C%5C-1c_%7B1%7D%26-4c_%7B1%7D%5Cend%7Barray%7D%5Cright%5D%20%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dc_%7B2%7D%260%5C%5C0%26c_%7B2%7D%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0%260%5C%5C0%260%5Cend%7Barray%7D%5Cright%5D)
And the system of equations is:
There are several methods to solve this system. One of them is to multiply the second equation to -1 and add both equations:
With , substitute in one of the equations and find :
<u>For the equation, </u><u> and </u><u />
