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3 years ago

1.8×10^-2-3.9×10^-3 what is the answer

Mathematics

2 answers:

Alja [10]3 years ago

3 0

The answer is 0.0141

KonstantinChe [14]3 years ago

3 0

1.8×10^-2 = 0.018

3.9×10^-3 = 0.0039

1.8×10^-2-3.9×10^-3

= 0.018 - 0.0039

= -0.0141

<span>Answer in scientific form notation:</span>

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3 years ago

Can some help me with this .

PtichkaEL [24]

Answer:

1. 169π ft²

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3) 25π yd²

Step-by-step explanation:

1) π * 13²

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169π ft²

2) π * 12²

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units = in²

144π in²

3)π * 5²

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2 years ago

Hannah has a aquarium with 25 Liter of water in it.How many milliliters of water are in the aquarium

Vikentia [17]

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25000

Step-by-step explanation:

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8 0

2 years ago

Read 2 more answers

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

The population p (in millions) of Italy from 1990 through 2008 can be approximated by the model p=56.8e^o.oo15t , where t repres

dimulka [17.4K]

As we already have the model that describes the change of the population in Italy in terms of the years that have elapsed, we only have to replace the conditions that are requested in that equation.

Therefore to find the population of Italy in the year 2000 (t = 10 years) substitute t = 10 in the equation and have:

$p = 56.8e^{0.0015*10}\\$