The legs are equal (congruent) since its isosceles the sides are equal and their base angles are also equal.
5/54 or approximately 0.092592593
There are 6^3 = 216 possible outcomes of rolling these 3 dice. Let's count the number of possible rolls that meet the criteria b < y < r, manually.
r = 1 or 2 is obviously impossible. So let's look at r = 3 through 6.
r = 3, y = 2, b = 1 is the only possibility for r=3. So n = 1
r = 4, y = 3, b = {1,2}, so n = 1 + 2 = 3
r = 4, y = 2, b = 1, so n = 3 + 1 = 4
r = 5, y = 4, b = {1,2,3}, so n = 4 + 3 = 7
r = 5, y = 3, b = {1,2}, so n = 7 + 2 = 9
r = 5, y = 2, b = 1, so n = 9 + 1 = 10
And I see a pattern, for the most restrictive r, there is 1 possibility. For the next most restrictive, there's 2+1 = 3 possibilities. Then the next one is 3+2+1
= 6 possibilities. So for r = 6, there should be 4+3+2+1 = 10 possibilities.
Let's see
r = 6, y = 5, b = {4,3,2,1}, so n = 10 + 4 = 14
r = 6, y = 4, b = {3,2,1}, so n = 14 + 3 = 17
r = 6, y = 3, b = {2,1}, so n = 17 + 2 = 19
r = 6, y = 2, b = 1, so n = 19 + 1 = 20
And the pattern holds. So there are 20 possible rolls that meet the desired criteria out of 216 possible rolls. So 20/216 = 5/54.
Complete Question:
a) Is it plausible that X is normally distributed?
b) For a random sample of 50 such pairs, what is the (approximate) probability that the sample mean courtship time is between 100 min and 125 min?
Answer:
a) It is plausible that X is normally distributed
b) probability that the sample mean courtship time is between 100 min and 125 min is 0.5269
Step-by-step explanation:
a)X denotes the courtship time for the scorpion flies which indicates that is a real - valued random variable, and since normal distribution is a continuous probability distribution for a real valued random variable, it is plausible that X is normally distributed.
b) Probability that the sample mean courtship time is between 100 min and 125 min
From the probability distribution table:
Huw's answer of 9 days is incorrect because more guests would finish the food in less time so the number of days for 30 guests is less than 6 days.
The number of days that the food can last 30 guests is 4 days.
<h3>How long will Seaview Hotel's food feed 30 guests?</h3>
More guests would eat the food faster which means that it would be finished in less days than the 6 days used by 20 guests.
To find the days the food would last 30 guests, use the inverse proportion formula:
= (Number of guests x Number of days food lasts for guests) / New number of guests
= (20 x 6) / 30
= 4 days
Full question is:
The only food provided for guests at Seaview Hotel is breakfast. The hotel has enough food to make breakfast for 20 guests for 6 days. How long would the food last 30 guests? You may assume each guest eats the same amount of food for breakfast.
Without working out the correct answer, explain why Huw’s answer of 9 days is incorrect.
Find out more on inverse proportion at brainly.com/question/1266676.
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