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ira [324]
4 years ago
12

What is 5.8616 ÷100

Mathematics
2 answers:
Effectus [21]4 years ago
6 0
5.8616÷100=
the answer is 0.05616
svetoff [14.1K]4 years ago
4 0
The answer of 5.8616 / 100 would be 0.058616. all you do is move the decimal 2 places to the left.
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(08.06)
Greeley [361]

Answer:

(x-7)(x+7)

Step-by-step explanation:

x^2 -49

This is the difference of squares

x^2 -7^2

We know that a^2 -b^2 = (a-b)(a+b)

(x-7)(x+7)

6 0
4 years ago
The length of a rectangle is 3 meters more than the width. The perimeter of the rectangle is 26 meters.
Maksim231197 [3]

l=b+3

perimeter=26

2(l+b)=26

l+b=13

b+3+b=13

2b=10

b=5m

l=b+3

= 5+3

=8m

8 0
3 years ago
If DK PVX, which of the following statements must also be true Check all that apply
Oxana [17]
I think it’s AVPXADK
8 0
3 years ago
Pls help I have doubts
azamat

Answer:

B

Step-by-step explanation:

See when you multiply 3 x 4 or 4 x 3 the answer is the same (is equivalent)

Similarly, -3 x 3 x -4/9 or 3 x -3 x 4/9 is equivalent

Therefore option(B)

6 0
3 years ago
Read 2 more answers
A certain hereditary disease can be passed from a mother to her children. Given that the mother has the disease, her children in
uranmaximum [27]

Answer:

(a) P=0.694

(b) It is independent, beacuse the probability of having the disease for the children depends only on her mother condition (if she has the disease or not), not the condition of his brothers and sisters.

(c) P=0.25

Step-by-step explanation:

(a) If the mother has 0.33 probabilities of having the disease, the probability of the children having the disease is equal to the product of the probability of the mother having the disease (0.33) and the probability of inherit it (0.50).

So the probability of one child of having the disease is 0.33*0.5=0.167. The probability of not having the disease is then (1-0.167)=0.833

The probability of both children to not have the disease is 0.833^2=0.694.

(b) It is independent, beacuse the probability of having the disease for the children depends only on her mother condition (if she has the disease or not), not the ones of his brothers and sisters.

(c) If the mother has the disease, the child have a probability of 0.5 of having the disease.

The probability, given that the mother has the disease, of both child not having the disease is 0.5^2=0.25.

4 0
4 years ago
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