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:
Step-by-step explanation:
![5\sqrt[3]{x+8}=35\\ \\ \sqrt[3]{x+8}=7\\ \\ x+8=7^3\\ \\ x+8=343](https://tex.z-dn.net/?f=5%5Csqrt%5B3%5D%7Bx%2B8%7D%3D35%5C%5C%20%5C%5C%20%5Csqrt%5B3%5D%7Bx%2B8%7D%3D7%5C%5C%20%5C%5C%20x%2B8%3D7%5E3%5C%5C%20%5C%5C%20x%2B8%3D343)
The answer to the question is 26°
180°-154°= 26°
8 hundred thousands and 1 thousand.
Answer:
3
Step-by-step explanation:
-2+1= -1
4-3=1
-3/1= -3