A) What is the lower quartile of the data represented in the following box-and-whisker plot?

The function w=30-3/4n represents the weight w in pounds of the papers in tyrones newspaper delivery bag after he delivers n new

adoni [48]

Answer:

domain (−∞,∞)

range (−∞,∞)

n = 40 and it refers to the fact that after delivering 40 newspapers, there will be no paper left in the bag

Step-by-step explanation:

To find the zero, we equate the linear equation to zero

w = 30 - 3n/4

3n/4 = 30

3n = 120

n = 120/3

n = 40

What this mean in this context is that the the bag becomes empty after he has delivered 40 newspapers or we can say that the maximum number of papers the bag can take is 40

The range refers to the number on the y-axis

Here the range is (−∞,∞)

The domain refers to the x axis values

We have (−∞,∞)

4 0

3 years ago

PLEASE ANSWER ASAP Which function has a constant additive rate of change of –1/4?

Hatshy [7]

Answer:

the first graph

Step-by-step explanation:

The first one - y = 1/4x + 3/2.

sorry im a bit late but hopefully this helps

6 0

2 years ago

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar

Oksana_A [137]

Answer:

$n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}$

Step-by-step explanation:

Lets divide it in cases, then sum everything

Case (1): All 5 numbers are different

In this case, the problem is reduced to count the number of subsets of cardinality 5 from a set of cardinality n. The order doesnt matter because once we have two different sets, we can order them descendently, and we obtain two different 5-tuples in decreasing order.

The total cardinality of this case therefore is the Combinatorial number of n with 5, in other words, the total amount of possibilities to pick 5 elements from a set of n.

${n \choose 5 } = \frac{n!}{5!(n-5)!}$

Case (2): 4 numbers are different

We start this case similarly to the previous one, we count how many subsets of 4 elements we can form from a set of n elements. The answer is the combinatorial number of n with 4 ${n \choose 4} .$

We still have to localize the other element, that forcibly, is one of the four chosen. Therefore, the total amount of possibilities for this case is multiplied by those 4 options.

The total cardinality of this case is $4 * {n \choose 4} .$

Case (3): 3 numbers are different

As we did before, we pick 3 elements from a set of n. The amount of possibilities is ${n \choose 3} .$

Then, we need to define the other 2 numbers. They can be the same number, in which case we have 3 possibilities, or they can be 2 different ones, in which case we have ${3 \choose 2 } = 3$ possibilities. Therefore, we have a total of 6 possibilities to define the other 2 numbers. That multiplies by 6 the total of cases for this part, giving a total of $6 * {n \choose 3}$

Case (4): 2 numbers are different

We pick 2 numbers from a set of n, with a total of ${n \choose 2}$ possibilities. We have 4 options to define the other 3 numbers, they can all three of them be equal to the biggest number, there can be 2 equal to the biggest number and 1 to the smallest one, there can be 1 equal to the biggest number and 2 to the smallest one, and they can all three of them be equal to the smallest number.