Response:
The probability that a student has a pet, given they do not have any siblings is:
Option: D ( 60%)
Step-by-step breakdown:
Let A represent the situation where a student lacks a sibling.
Let B signify the occurrence that a student has a pet.
Consequently, A∩B refers to the event in which a student is without siblings but possesses a pet.
Let P denote the chance of an event happening.
We need to determine:
P(B|A)
From our knowledge:
From the data provided:
P(A)=0.25
and P(A∩B)=0.15
Thus,
which expressed as a percentage is:
Therefore, the probability is:
60%
1
2
3
Step-by-step explanation: Generally, during the roll of two fair 6-sided dice, the doubles result in (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6). Therefore, the total for doubles is N = 6. The outcome of rolling two fair 6-sided dice yields n = 36. Thus, the probability of rolling doubles (matching numbers on both dice) is calculated mathematically. When rolling two fair dice, outcomes that sum to 4 or less are (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1). Observing this, we see two doubles present. Consequently, the conditional probability of rolling doubles is represented mathematically. Lastly, when rolling the two fair dice, outcomes that show different numbers result in L = 30, while outcomes where at least one die shows a 1 give W = 10. Hence, the conditional probability of having at least one die show a 1 is presented mathematically.
Answer:
A relationship exists between age and duration of stay; however, causation has not yet been established. Additional research would be necessary to clarify this matter.
Step-by-step explanation:
The answer was checked
Honestly, I find Mrs. Garcia's method easier to perform mentally. It hinges on how familiar you are with your multiples of 5. (5*15 = 75 is a multiplication I often use)
Melissa's approach involves calculating 5*20 = 100 and 5*9 = 45, then combines the 3-digit result 100 with the 2-digit result 45, yielding 145. Adding 45 to 00 is simple and doesn’t require carrying digits, thus the arithmetic is fairly straightforward.
Mrs. Garcia's technique involves computing 5*14 = 70 and 5*15 = 75, then summing these two-digit results. Many people may not readily recall that 5*15=75, which complicates forming that product. The addition of 70 and 75 requires a carrying operation, making the math somewhat more complex. The resulting total is 145.
(The rationale behind my preference for Mrs. Garcia's method is that I can achieve the final sum by simply doubling 7 tens, followed by adding 5. The only 3-digit number to remember mentally is the ultimate total.)
_____Subtraction introduces a slight complication, yet reshaping it as $5(30 -1) = $150 - 5 = $145 is possible.
Or, you may reframe it as $5(28 +1) = $140 +5 = $145.
Dividing an even number by 2 to find the product of 5 is straightforward when you append a zero.
5*14 = 10*7 = 70
5*28 = 10*14 = 140.
