Answer:
3/6 and 5/10
Explanation:
Given the fraction 1/2:
(a)Multiply the numerator and denominator by 3
An equivalent fraction is 3/6.
(b)Multiply the numerator and denominator by 5:
An equivalent fraction is 5/10.
Answer:
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Step-by-step explanation:eeeeeeeeeeeeeeeee
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For this case , the parent function is given by [tex f (x) =x^2
[\tex]
We apply the following transformations
Vertical translations :
Suppose that k > 0
To graph y=f(x)+k, move the graph of k units upwards
For k=9
We have
[tex]h(x)=x^2+9
[\tex]
Horizontal translation
Suppose that h>0
To graph y=f(x-h) , move the graph of h units to the right
For h=4 we have :
[tex ] g (x) =(x-4) ^ 2+9
[\tex]
Answer :
The function g(x) is given by
G(x) =(x-4)2 +9
Answer:
probability that the other side is colored black if the upper side of the chosen card is colored red = 1/3
Step-by-step explanation:
First of all;
Let B1 be the event that the card with two red sides is selected
Let B2 be the event that the
card with two black sides is selected
Let B3 be the event that the card with one red side and one black side is
selected
Let A be the event that the upper side of the selected card (when put down on the ground)
is red.
Now, from the question;
P(B3) = ⅓
P(A|B3) = ½
P(B1) = ⅓
P(A|B1) = 1
P(B2) = ⅓
P(A|B2)) = 0
(P(B3) = ⅓
P(A|B3) = ½
Now, we want to find the probability that the other side is colored black if the upper side of the chosen card is colored red. This probability is; P(B3|A). Thus, from the Bayes’ formula, it follows that;
P(B3|A) = [P(B3)•P(A|B3)]/[(P(B1)•P(A|B1)) + (P(B2)•P(A|B2)) + (P(B3)•P(A|B3))]
Thus;
P(B3|A) = [⅓×½]/[(⅓×1) + (⅓•0) + (⅓×½)]
P(B3|A) = (1/6)/(⅓ + 0 + 1/6)
P(B3|A) = (1/6)/(1/2)
P(B3|A) = 1/3
