the relation described in this statement can be classified as which of the following: a function, a relation, both a function an
1 answer:
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See the picture attached to better understand the problem
we know that
in the right triangle ABC
cos A=AC/AB
cos A=1/3
so
1/3=AC/AB----->AB=3*AC-----> square----> AB²=9*AC²----> equation 1
applying the Pythagoras Theorem
BC²+AC²=AB²-----> 2²+AC²=AB²---> 4+AC²=AB²----> equation 2
substitute equation 1 in equation 2
4+AC²=9*AC²----> 8*AC²=4----> AC²=1/2----> AC=√2/2
so
AB²=9*AC²----> AB²=9*(√2/2)²----> AB=(3√2)/2
the answer is
the hypotenuse is (3√2)/2
we know that
in the right triangle ABC
cos A=AC/AB
cos A=1/3
so
1/3=AC/AB----->AB=3*AC-----> square----> AB²=9*AC²----> equation 1
applying the Pythagoras Theorem
BC²+AC²=AB²-----> 2²+AC²=AB²---> 4+AC²=AB²----> equation 2
substitute equation 1 in equation 2
4+AC²=9*AC²----> 8*AC²=4----> AC²=1/2----> AC=√2/2
so
AB²=9*AC²----> AB²=9*(√2/2)²----> AB=(3√2)/2
the answer is
the hypotenuse is (3√2)/2
Answer:
y 0 1 2 3
P(Y=y) 0.0676 0.3549 0.3875 0.19
Step-by-step explanation:
P(Wed) = 0.26
P(Thurs) = 0.39
P(Fri) = 0.25
P(Sat) = 0.10
Y = No. of days beyond Wednesday it takes for both magazines to arrive i.e. 0,1,2,3
Y=0 means the magazines will arrive on Wednesday
Y=1 means the magazines will arrive till Thursday
Y=2 means the magazines will arrive till Friday
Y=3 means the magazines will arrive till Saturday
The possible combinations for Y are
Y(W,W) Y(W,T) Y(W,F) Y(W,S)
Y(T,W) Y(T,T) Y(T,F) Y(T,S)
Y(F,W) Y(F,T) Y(F,F) Y(F,S)
Y(S,W) Y(S,T) Y(S,F) Y(S,S)
So, we can classify these possible outcomes as Y=0,1,2,3.
Y(0) = Y(W,W) (both magazines take 0 days to arrive beyond Wednesday)
Y(1) = Y(W,T), Y(T,T), Y(T,W) (both magazines take 1 day to arrive beyond Wednesday)
Y(2) = Y(W,F), Y(T,F), Y(F,F) Y(F,W) Y(F,T) (both magazines arrive till Friday)
Y(3) = Y(W,S), Y(T,S), Y(F,S), Y(S,W), Y(S,T), Y(S,F), Y(S,S) (both magazines arrive till Saturday)
To calculate the PMF, we need to calculate the probability for each of the points in Y(0,1,2,3).
Y(0) = Y(W,W)
= 0.26 x 0.26
Y(0) = 0.0676
Y(1) = Y(W,T) + Y(T,T) + Y(T,W)
= (0.26 x 0.39) + (0.39 x 0.39) + (0.39 x 0.26)
= 0.1014 + 0.1521 + 0.1014
Y(1) = 0.3549
Y(2) = Y(W,F) + Y(T,F) + Y(F,F) + Y(F,W) + Y(F,T)
=(0.26 x 0.25) + (0.39 x 0.25) + (0.25 x 0.25) + (0.25 x 0.26) + (0.25 x 0.39)
= 0.065 + 0.0975 + 0.0625 + 0.065 + 0.0975
Y(2) = 0.3875
Y(3) = Y(W,S) + Y(T,S) + Y(F,S) + Y(S,W) + Y(S,T) + Y(S,F) + Y(S,S)
= (0.26 x 0.10) + (0.39 x 0.10) + (0.25 x 0.10) + (0.10 x 0.26) + (0.10 x 0.39) + (0.10 x 0.25) + (0.10 x 0.10)
= 0.026 + 0.039 + 0.025 + 0.026 + 0.039 + 0.025 + 0.010
Y(3) = 0.19
y 0 1 2 3
P(Y=y) 0.0676 0.3549 0.3875 0.19
The PMF plot is attached as a photo here.
