Answer:
1.92%
Step-by-step explanation:
The probability for first case, picking a queen out of deck, will be:
as there will be 4 queens in a deck, one of each suit.
For the second pick, the probability of picking a diamond card, will be:
here the total will remain 52 as he has replaced the first card and not kept it aside and there will be 13 cards in diamond suit (including the three face cards).
Thus the net probability for both cases will be:
Thus total probability for the combined two cases will be 1.92%
X^2+x=2
first set to zero
x^2+x=2
subtract 2 from both sides
x^2+x-2=0
factor because if you have
x times y=0 then assume x and/or y=0 so
factor x^2+1x-2
to factor you find what 2 number add to 1 and multiply to get -2
the numbers are -1 and 2 so
it factors out to
(x-1)(x+2)=0
assume each is zero
x-1=0
add 1 to both sides
x=1
x+2=0
subtract 2 from both sides
x=-2
x=1 or -2
Answer:
The answer is below
Step-by-step explanation:
A linear graph has an equation of the form:
y = mx + b,
where y and x are variables, m is the slope (rate of change) of the graph and b is the y intercept (value of y when x is 0).
Given that:
2x -2y = 4 , 3x + 2y = 6
The matrix form of the equation is:
AX = B
![A=\left[\begin{array}{cc}2&-2\\3&2\end{array}\right] ,X=\left[\begin{array}{c}x\\y\end{array}\right] ,B=\left[\begin{array}{c}4\\6\end{array}\right] \\\\Therefore:\\\\\left[\begin{array}{cc}2&-2\\3&2\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] =\left[\begin{array}{c}4\\6\end{array}\right] \\\\\left[\begin{array}{c}x\\y\end{array}\right] =\left[\begin{array}{cc}2&-2\\3&2\end{array}\right]^{-1} \left[\begin{array}{c}4\\6\end{array}\right] \\\\](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%26-2%5C%5C3%262%5Cend%7Barray%7D%5Cright%5D%20%2CX%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%2CB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%5C%5C6%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5CTherefore%3A%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%26-2%5C%5C3%262%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%5C%5C6%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%26-2%5C%5C3%262%5Cend%7Barray%7D%5Cright%5D%5E%7B-1%7D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%5C%5C6%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C)
![\left[\begin{array}{c}x\\y\end{array}\right] =\left[\begin{array}{cc}0.2&0.2\\-0.3&0.2\end{array}\right] \left[\begin{array}{c}4\\6\end{array}\right] \\\\\left[\begin{array}{c}x\\y\end{array}\right] =\left[\begin{array}{c}2\\0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.2%260.2%5C%5C-0.3%260.2%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%5C%5C6%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
We have a right triangle with
a = one leg = 28
b = other leg = 21
c = hypotenuse = 5x
use the pythagorean theorem (a^2+b^2 = c^2) and solve for x
a^2+b^2 = c^2
28^2+21^2 = (5x)^2
784+441 = 25x^2
1225 = 25x^2
25x^2 = 1225
(25x^2)/25 = 1225/25
x^2 = 49
sqrt(x^2) = sqrt(49)
|x| = 7
x = 7 or x = -7
Since the hypotenuse has length 5x meters, this means
5x = 5*7 = 35
or
5x = 5(-7) = -35
Toss out the negative length as it makes no sense.
The only value of x is x = 7
Therefore the final answer is x = 7
The median of a set of numbers is the midpoint of the set if you were to organize the numbers from least to greatest or greatest to least. So, all we have to do is rearrange the numbers from least to greatest and find the midpoint:
67 76 76 82 84 87 93
count 3 in from both sides and the remaining number is 82, which is the median.
