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At a bake sell, 3/5 of the baked goods are pies. the rest of the baked goods are the plates of cookies. there are 24 plates of c
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Answer:
a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.
b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.
c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.
Step-by-step explanation:
There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.
Binomial probability
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 combinatios of x objects from a set of n elements, given by the following formula.
And is the probability of X happening.
In this problem, we have that:
10 student are sampled, so
34% of college students say they use credit cards because of the rewards program, so
(a) exactly two
This is P(X = 2).
There is a 18.73% probability that exactly two students use credit cards because of the rewards program.
(b) more than two
This is .
Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:
In which
So
There is a 71.62% probability that more than two students use credit cards because of the rewards program.
(c) between two and five inclusive
This is:
There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.
Answer:
1. (x - 3)² = 8
2. (x + 2)² = 3
3. (x + 6)² =
4. (x + 3)² = 27
5. (x + 4)² = 13
6.
Step-by-step explanation:
Completion of Square:
In the following problems the terms in the RHS of the above equation may be missing. We balance the equation. Simplify it and re write it in terms of LHS.
1. x² - 6x + 1 = 0
Taking the constant term to the other side, we get:
x² - 6x = - 1
⇒ x² - 2(3)x = -1
⇒ x² -2(3)x + 9 = - 1 + 9 [Adding 9 to both the sides]
⇒ x² -2(3)x + 3² = 8
⇒ (x - 3)² = 8 is the answer.
2. 3x² + 12x + 3 = 0
Note that the co-effecient of x² is not 1. We make it 1, by dividing the whole equation by 3. And then proceed like the previous problem.
3x² + 12x = -3
Dividing by 3 through out, x² + 4x = - 1
⇒ x² + 2(2) + 4 = -1 + 4
⇒ x² +2(2) + 2² = 3
⇒ (x + 2)² = 3 is the answer.
3. 2x² + 24x = 29
x² + 12x =
⇒ x² + 2(6)x + 36 = + 36
⇒ x² + 2(6)x + 6² =
⇒ (x + 6)² = is the answer.
4. x² + 6x - 18 = 0
x² + 6x = 18
⇒ x² + 2(3)x = 18
⇒ x² + 2(3)x + 9 = 18 + 9
⇒ x² + 2(3)x + 3² = 27
⇒ (x + 3)² = 27 is the answer.
5. x² + 8x + 3 = 0
x² + 8x = -3
⇒ x² + 2(4)x = -3
⇒ x² + 2(4)x + 16 = - 3 + 16
⇒ x² + 2(4)x + 16 = 13
⇒ (x + 4)² = 13 is the answer.
6. 9x² - 30x + 6 = 0
9x² - 30x = - 6
⇒ x² x = - 6
is the answer.