Answer:
The probability that the red card came from the first stack is:
40%
Step-by-step explanation:
a) Data and Calculations:
Stack Red Blue Total
First 6 5 11
Second 9 0 9
Total 15 5 20
The probability of selecting from the first stack = 50% since equal opportunity is given to the two sacks.
The probability of red card from the first stack = 6/15 = 0.4 = 40%.
b) The probability that the red card came from the first stack is given by the number of red cards in the first stack divided by the number or red cards in the two stacks. This is equal to 6/15 = 0.4 or 40%.
Answer:
Step-by-step explanation:
The formula for the length of a vector/line in your case.
![L = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = \sqrt{[4 - (-1)]^2 + [2 -(-3)]^2} = \sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2}](https://tex.z-dn.net/?f=L%20%3D%20%5Csqrt%7B%28x_2-x_1%29%5E2%20%2B%20%28y_2-y_1%29%5E2%7D%20%3D%20%5Csqrt%7B%5B4%20-%20%28-1%29%5D%5E2%20%2B%20%5B2%20-%28-3%29%5D%5E2%7D%20%3D%20%5Csqrt%7B5%5E2%20%2B%205%5E2%7D%20%3D%20%5Csqrt%7B50%7D%20%3D%205%5Csqrt%7B2%7D)
Your answer is going to be fifteen plus r
Answer:
x = 6
y = 4
Step-by-step explanation:
Let the two numbers be x and y
<u><em>Condition 1:</em></u>
7x+3y = 54 -----------(1)
<u><em>Condition 2:</em></u>
x = 2+y -----------------(2)
<em>Putting (2) in (1)</em>
=> 7(y+2)+3y = 54
=> 7y+14+3y = 54
=> 10y = 54-14
=> 10y = 40
<em>Dividing both sides by 10</em>
=> y = 4
<em>Now putting y = 4 in eq(2)</em>
=> x = 2+4
=> x = 6
