Step-by-step explanation:
Area of circular pond=22/7×14×14
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>2</em><em>2</em><em>/</em><em>7</em><em>×</em><em>1</em><em>9</em><em>6</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>6</em><em>1</em><em>6</em><em>m</em>
If we draw the contingency table of x (vertical) against y (horiz.), we have a square.
For n=4, we have (legend: < : x<y = : x=y > : x>y
y 1 2 3 4
x
1 = < < <
2 > = < <
3 > > = <
4 > > > =
We see that there are n(n-1)/2 cases of x<y out of n^2.
Therefore,
p(x<y)=n(n-1)/(2n^2)=(n-1)/(2n)
However, if the sample space is continuous, it will be simply p(x<y)=1/2.
Answer:
50=(3x)x
Step-by-step explanation:
50=(3x)x. The question is only asking for the inequality of the problem. Since the area of a rectangle is L x W, you would write L as 3x and W as x, and since the area of the rectangle is given, you can set the area as 50.
First we need to know how much each pen costs. So we take 11.77 and divide it by 11. That gives us 1.07 per pen. So answer choice A has 4 pens for 4.44, so we need to multiply 1.07 x 4. That gives you 4.28, so we know that isn't the correct answer. For option B, we need to multiply 1.07 x 5. That is 5.35, so that also can't be the answer. For C, we need to do 1.07 x 6, and that gives us 6.42. So we know the correct answer is option C.
*The answer is A. 24.*
All you have to do is back solve. In other words, just plug your answer choices into the equation.