Answer:
210
Step-by-step explanation:
Here comes the problem from Combination.
We are being asked to find the number of ways out in which 3 students may sit on 7 seats in a row. Please see that in this case the even can not be repeated.
Let us start with the student one. For him all the 7 seats are available to sit. Hence number of ways for him to sit = 7
Let us see the student second. For him there are only 6 seats available to sit as one seat has already been occupied. Hence number of ways for him to sit = 6
Let us see the student third. For him there are only 5 seats available to sit as two seat has already been occupied. Hence number of ways for him to sit = 5
Hence the total number of ways for three students to be seated will be
7 x 6 x 5
=210
Expanded FormExpanded form refers to a base and an exponent written as repeated multiplication. ExponentExponents are used to describe the number of times that a term is multiplied by itself. ExpressionAn expression is a mathematical phrase containing variables, operations and/or numbers.
Answer:
3 3/8
Step-by-step explanation:
6 minus 2 5/8 equals 3 3/8
3 3/8 plus 2 5/8 equals 6
Answer:
Original price reduced by 30%
Step-by-step explanation:
You want the deepest discount, which means you want the greatest percent/amount off. The first choice is 20%. The second is 30%. The third one is 25%, because you are paying for 75% (100-75). The second choice offers the deepest discount because it is 30% off.
Using the hypergeometric distribution, it is found that there is a 0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.
In this problem, the bulbs are chosen without replacement, hence the <em>hypergeometric distribution</em> is used to solve this question.
<h3>What is the hypergeometric distribution formula?</h3>
The formula is:
The parameters are:
- x is the number of successes.
- N is the size of the population.
- n is the size of the sample.
- k is the total number of desired outcomes.
In this problem:
- There are 12 bulbs, hence N = 12.
- 3 are defective, hence k = 3.
The third defective bulb is the fifth bulb if:
- Two of the first 4 bulbs are defective, which is P(X = 2) when n = 4.
- The fifth is defective, with probability of 1/8, as of the eight remaining bulbs, one will be defective.
Hence:
0.2182 x 1/8 = 0.0273.
0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.
More can be learned about the hypergeometric distribution at brainly.com/question/24826394
