Conditional probablility P(A/B) = P(A and B) / P(B). Here, A is sum of two dice being greater than or equal to 9 and B is at least one of the dice showing 6. Number of ways two dice faces can sum up to 9 = (3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6) = 10 ways. Number of ways that at least one of the dice must show 6 = (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6), (6, 5), (6, 4), (6, 3), (6, 2), (6, 1) = 11 ways. Number of ways of rolling a number greater than or equal to 9 and at least one of the dice showing 6 = (3, 6), (4, 6), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6) = 7 ways. Probability of rolling a number greater than or equal to 9 given that at least one of the dice must show a 6 = 7 / 11
Answer:
x=133
Step-by-step explanation:
Given: m║n
This means that 147 and x+14 are <em>alternate exterior angles</em>. This also means that they are equivalent. We can use this information to set up an equation:
x+14=147
Subtract 14 from both sides to isolate x
x=133
1. 2/3. Flip 2/3 into 3/2 and then multiply and simplify.
2. 30/91. Flip 7/6 to 6/7 and then multiply. You cannot simplify the fraction.
3. 26/27. Flip 9/10 to 10/9 and then multiply. You cannot simplify the fraction.
4. 44.2. Multiply it as if there was no decimal. Then count the number of digits after the decimal in each factor. Then put the same number of digits behind the decimal in the product.
5. 98.75. Multiply it as if there was no decimal. Then count the number of digits after the decimal in each factor. Then put the same number of digits behind the decimal in the product.
6. 3.36. Multiply it as if there was no decimal. Then count the number of digits after the decimal in each factor. Then put the same number of digits behind the decimal in the product.
7. 2. Multiply the divisor by as many 10’s as necessary until you get a whole number. Remember to multiply the dividend by the same number of 10’s. Then divide it normally.
8. 10.93 (rounded). Multiply the divisor by as many 10’s as necessary until you get a whole number. Remember to multiply the dividend by the same number of 10’s. Then divide it normally. I rounded it to the hundredth.
Hope this helps!
12/60 students chose science fiction
Approximately x/150 students prefer sf
60x = 1800
x = 30
30/150 students are assumed to prefer sf
30/150 = x/100
150x = 3000
x = 20
20/100 students are likely to prefer sf
Mr. Rodriguez made a reasonable estimate for the approximate percentage of students that prefer science fiction, because if 12/60 is equivalent to 30/150 which refers to the number of students who can be assumed to prefer science Fiction out of the whole school. Considering we need to identify what 30/150 as a percentage is, we can reduce it down to 1/5 to make I easier, then divide 1 by 5 to get .2
.2 as a percentage is 20%, so his inference was indeed reasonable.
(♥ω♥*)Brainliest Please(♥ω♥*)
