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2 years ago

7

In previous months, you ordered 80 boxes of pens per month. This month, you want to decrease the number of boxes of pens you ord

er by 40%. How many boxes should you order this month?

2 answers:

hoa [83]2 years ago

8 0

If you have 80 boxes then 40% of boxes is the number 32 , 80-32=48 so you would order 48 boxes

Lapatulllka [165]2 years ago

3 0

80(0.40)=32
80-32=48 boxes of pens

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Tresset [83]

Answer:

1)5873 x 4=23492

2)5000 x 4=20000

3)800 x 4=3200

4)70 x 4=280

5)3 x 4=12

Step-by-step explanation:

4 0

2 years ago

Roberta needs 12 zippers to make 3 of the same backpacks. If Roberta has 48 zippers, how many of those same backpacks can she ma

yaroslaw [1]

First, you need to find out how many zippers are used to make one bookbag. To do that, you divide the zippers used to make the 3 backpacks.

12 / 3 = 4

So, 4 zippers are used per bookbag. Then, you divide the number of available zippers by four to find out how many bags she can make.

48 / 4 = 12

Roberta can make 12 bookbags.

5 0

3 years ago

14×12 distributive property

miss Akunina [59]

186 if that’s the answer you where looking for if not comment here it will give me a notification and I will see

5 0

3 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.

The total amount of possibilities for this case is

$4 * {n \choose 2}$

Case (5): All numbers are the same

This is easy, he have as many possibilities as numbers the set has. In other words, n

Conclussion

By summing over all 5 cases, the total amount of possibilities to form 5-tuples of integers from 1 through n is

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

I hope that works for you!

4 0

3 years ago

A university knows from historical data that 25% of students in an introductory statistics class withdraw before completing the

Vikki [24]

Answer:

13.4%

Step-by-step explanation:

Use binomial probability:

P = nCr p^r q^(n-r)

where n is the number of trials,

r is the number of successes,

p is the probability of success,

and q is the probability of failure (1-p).

Here, n = 16, r = 2, p = 0.25, and q = 0.75.

P = ₁₆C₂ (0.25)² (0.75)¹⁶⁻²

P = 120 (0.25)² (0.75)¹⁴

P = 0.134

There is a 13.4% probability that exactly 2 students will withdraw.

8 0

3 years ago