A Situation for y=5x
1 answer:
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Ok , lots of questions=lower standard of explanation
just answers
5. 8.5 times 10^12
6. 0.001260-7,003,000
7.3.843 times 10^4
9. x=1
10. 7^(x+5)=49^(x+3)
11. 1.03
12. 25,700(0.85)^x
13. 10a√a
14.![2 x^{2} y^{4} \sqrt[3]{5x^{2}}](https://tex.z-dn.net/?f=2%20x%5E%7B2%7D%20%20y%5E%7B4%7D%20%20%5Csqrt%5B3%5D%7B5x%5E%7B2%7D%7D%20)
15.6√3
16.
17.![4 \sqrt[3]{x^{2}}](https://tex.z-dn.net/?f=%204%20%5Csqrt%5B3%5D%7Bx%5E%7B2%7D%7D%20)
18.
19.√109
20. (8,(3/2))
21. 5 units
just answers
5. 8.5 times 10^12
6. 0.001260-7,003,000
7.3.843 times 10^4
9. x=1
10. 7^(x+5)=49^(x+3)
11. 1.03
12. 25,700(0.85)^x
13. 10a√a
14.
15.6√3
16.
17.
18.
19.√109
20. (8,(3/2))
21. 5 units
Answer:
0.2773 = 27.73% probability that at the May celebration, exactly two members of the group have May birthdays
Step-by-step explanation:
For each person, there are only two possible outcomes. Either they have a birthday in May, or they do not. The probability of a person having a birthday in May is independent of any other person. This means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
Probability of a person being in May:
May has 31 days in a year of 365. So
Group of 20 friends:
This means that
What is the probability that at the May celebration, exactly two members of the group have May birthdays?
This is P(X = 2).
0.2773 = 27.73% probability that at the May celebration, exactly two members of the group have May birthdays
You need to solve for x or y in one equation and and substitute it into the other equation in place of the variable that corresponds to it.
