Answer:
The chance that there will be exactly 2 heads among the first five tosses and exactly 4 heads among the last 5 tosses is P=0.0488.
Step-by-step explanation:
To solve this problem we divide the tossing in two: the first 5 tosses and the last 5 tosses.
Both heads and tails have an individual probability p=0.5.
Then, both group of five tosses have the same binomial distribution: n=5, p=0.5.
The probability that k heads are in the sample is:

Then, the probability that exactly 2 heads are among the first five tosses can be calculated as:

For the last five tosses, the probability that are exactly 4 heads is:

Then, the probability that there will be exactly 2 heads among the first five tosses and exactly 4 heads among the last 5 tosses can be calculated multypling the probabilities of these two independent events:

Answer: 45 tomatillos
Step-by-step explanation: 30 tomatillos divided by 8 ounces is 3.75 tomatillos per ounce. So, multiply 3.75 by 12 ounces. What do we get? 45 tomatillos!
Yes all the water will fit and more because to find a volume of an object you multiple the height x length x width and in this case when all of it is calculated the base could hold 800 cm3 of water