Which ordered pair is a solution to the following:
1 answer:
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Answer:
Probability of rolling doubles in first three roles = 0.4124
Probability of it taking between 5 to 7 rolls = 0.1988
Step-by-step explanation:
Let's start by the total number of possible outcomes. This is 6 for one dice, and 6 for the other. So in total that's 6 x 6 = 36 different outcomes of the dice roll.
The number of ways we can roll a double is 6.
So the probability of rolling a double would be the number of ways to roll a double divided by the total possible outcomes.
This is : = 0.1667
The probability of NOT rolling a double: 1 - 0.1667 = 0.8333
Since the probability of rolling a double is equal to 1 - probability of not rolling a double, we can answer the first question in the following way.
The probability of NOT rolling a double for the first THREE turns: 0.8333 x 0.8333 x 0.8333 = 0.5786
Probability of rolling a double in first THREE turns: 1 - 0.5786 = <u>0.4124</u>
The probability it will take between 5 to 7 rolls would be:
EQUATION 1: (probability of rolling no doubles in first 4 turns) x (probability of rolling a double in the next three turns)
Since we have already found the probability of rolling a double in three turns to be 0.4124, all we need to find is the probability of NOT rolling a double in the first four turns.
This is: 0.8333 * 0.8333 * 0.8333 * 0.8333 = 0.4822
Plugging this into EQUATION 1 we get: 0.4822 * 0.4124 = <u>0.1988</u>
Answer:
13 inches
Step-by-step explanation:
Let P(t) and P(s) be the perimeters of the equilateral triangle and square respectively. Similarly, t and s be the side lengths of equilateral triangle and square respectively.
According to the first condition :
(Perimeter of an equilateral triangle is 11 inches more than the perimeter of a square)
According to the second condition :
(The side of the triangle is 6 inches longer than the side of the square)
From equations (1) & (2)
3(s + 6) = 4s + 11
3s + 18 = 4s + 11
3s - 4s = 11 - 18
-s = - 7
s = 7 inches
Thus the side of the triangle is 13 inches long.