The central 50 percent of data around average
Answer: No es posible (It is not possible)
Step-by-step explanation:
The question in english is as follows:
Is it possible to draw 9 line segments so that each segment crosses exactly 1 of the other segments?
A necessary condition for two lines to intersect is that they must be in the same plane (they must be coplanar), being their intersection a single point.
Now, if we want several lines to intersect only once, we need to have an <u>even number</u> of line segments. However, this is not the case because <u>9 is odd.
</u>
Therefore, it is not possible.
Y = yearbook club
d = drama club
y + d = 84
d = y + 16
now we sub y + 16 in for d in the first equation
y + d = 84
y + (y + 16) = 84
2y + 16 = 84
2y = 84 - 16
2y = 68
y = 68/2
y = 34 <=== yearbook club students
d = y + 16
d = 34 + 16
d = 50 <=== drama club students
I'm assuming you meant to say
P(A) = 2/3
P(A and B) = 1/3
If that is the case, then A and B are independent if and only if the following equation is true
P(A and B) = P(A)*P(B)
So we multiply P(A) and P(B) to get the value of P(A and B). We don't know what P(B) is, but we can use algebra to find it
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P(A and B) = P(A)*P(B)
P(A)*P(B) = P(A and B)
(2/3)*P(B) = 1/3
P(B) = (1/3)*(3/2) .... multiply both sides by the reciprocal of 2/3
P(B) = (1*3)/(3*2)
<h3>P(B) = 1/2 is the answer</h3>
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If P(B) = 1/2, then
P(A and B) = P(A)*P(B)
P(A and B) = (2/3)*(1/2)
P(A and B) = (2*1)/(3*2)
P(A and B) = 1/3
Which is the given probability for both events happening. This confirms we have the correct P(B) value.
11 5/8 in = 11.625 in
9 3/4in = 9.75 in
10 1/2 in= 10.5 in
11.625+9.75+10.5= 31.875in
To find the average divide the total of inches by the amount of pieces there are to find the average:
31.875 / 3 = 10.625in
