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2 years ago

Tai’s baby brother weighs 3,745 grams. What is his weight in kilograms?

Mathematics

2 answers:

jeyben [28]2 years ago

4 0

Answer:

3.745 kilograms

Step-by-step explanation:

I used a gram to kilograms calculator

Rus_ich [418]2 years ago

3 0

Answer:

3.745 kg

Step-by-step explanation:

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During a sale, CDs that normally cost $6.98 each were priced at 2 for $12.50. Pete bought 4 CDs at the sale price. How much mone

RSB [31]

Answer:

2.92

Step-by-step explanation:

Normal price

4 * 6.98 =27.92

Sale price

2 for 12.50 means 4 at 2* 12.50

2*12.50 = 25

Subtract

27.92-25=2.92

He saved 2.92

4 0

3 years ago

F(x)=4x squared −7x+7, Find f(2)

Lana71 [14]

Answer: 9

Step-by-step explanation:

4x²-7x+7 f(2)=4*(2)^2-7(2)+7=16-14+7=9

7 0

2 years ago

Solve triangle ABC, when A= 6, B=10 and c=12

valentinak56 [21]

Use cosine rule,

cos(A)=(b^2+c^2-a^2)/(2bc)
=(10^2+12^2-6^2)/(2*10*12)
=13/15
A=29.926 degrees.................................(A)

cos(B)=(c^2+a^2-b^2)/(2ca)
=(12^2+6^2-10^2)/(2*12*6)
=5/9
B=56.251 degrees.................................(B)

cos(C)=(a^2+b^2-c^2)/(2ab)
=(6^2+10^2-12^2)/(2*6*10)
=-1/15
C=93.823 degrees.................................(C)

Check:29.926+56.251+93.823=180.0 degrees....ok

5 0

3 years ago

X/x+2 = 4/5. Solve for x.

gregori [183]

4 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