Resuelve con calculadora y escribe la regla utilizada
1 answer:
When you multiply by 10, 100, or 1000, you must move the decimal place over by how many zeroes there are. Once there is no more decimals, you add 0's to the end of the number until you are done.
For example, 5.6 x 1000
1000 has 3 zeroes.
That means we need to move the decimal spot 3 times.
5.6 x 1000 = 56 x 100
We got rid of the decimal but we still have 2 zeroes left. So now we just add 2 zeroes at the end of the number.
5.6 x 1000 = 56 x 100 = 5600
I hope that helps! :)
You might be interested in
Answer:
1. The Venn diagrams are attached
2. When the statistics students number = 10·x + 3, we have;
The number of students that study
a. Algebra = 128/11
b. Statistic = 213/11
When the statistics students number = 2·x + 3, we have;
The number of students that study
a. Algebra = 16
b. Statistic = 15
Step-by-step explanation:
The parameters given are;
Total number of students = 30
Number of students that study algebra n(A) = x + 10
Number of students that study statistics n(B) = 10·x + 3
Number of student that study both algebra and statistics n(A∩B) = 4
Number of student that study only algebra n(A\B) = 2·x
Number of students that study neither algebra or statistics n(A∪B)' = 3
Therefore;
The number of students that study either algebra or statistics = n(A∪B)
From set theory we have;
n(A∪B) = n(A) + n(B) - n(A∩B)
n(A∪B) = 30 - 3 = 27
Therefore, we have;
n(A∪B) = x + 10 + 10·x + 3 - 4 = 27
11·x+13 = 27 + 4 = 31
11·x = 18
x = 18/11
The number of students that study
a. Algebra
n(A) = 18/11 + 10 = 128/11
b. Statistic
n(B) = 213/11
Hence, we have;
n(A - B) = n(A) - n(A∩B) = 128/11 - 4 = 84/11
Similarly, we have;
n(B - A) = n(B) - n(A∩B) = 213/11 - 4 = 169/11
However, assuming n(B) = (2·x + 3), we have;
n(A∪B) = n(A) + n(B) - n(A∩B)
n(A∪B) = 30 - 3 = 27
Therefore, we have;
n(A∪B) = x + 10 + 2·x + 3 - 4 = 27
2·x+3 + x + 10= 27 + 4 = 31
3·x = 18
x = 6
Therefore, the number of students that study
a. Algebra
n(A) = 16
b. Statistics
n(B) = 15
Hence, we have;
n(A - B) = n(A) - n(A∩B) = 16 - 4 = 12
Similarly, we have;
n(B - A) = n(B) - n(A∩B) = 15 - 4 = 11
The Venn diagrams can be presented as follows;
