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seropon [69]
3 years ago
7

Find the prime facrorization for 100 A.2² × 5² B.2×50 C.4×25 D.2³ × 5² E. 2² × 5³ ​

Mathematics
2 answers:
d1i1m1o1n [39]3 years ago
6 0

Answer:

A

Step-by-step explanation:

100-25*4=5 squared*2 sqaured

so A

sertanlavr [38]3 years ago
6 0

Answer:

prime factorization of 100=a.2²+5²

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Answer:

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Step-by-step explanation:

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3 years ago
Given that Segment BE is 5 units, find the indicated measures for each
zalisa [80]

Answer:

The indicated measure of the length of AE = 5 units

Step-by-step explanation:

To determine:

What is the indicated measure of length?

Information Fetching and Solution Steps:

From this single diagram, it seems the indicated measure of length which we have to determine is the length of AE.

Given that Segment BE is 5 units. i.e. BE = 5 units

  • As the square has two equal diagonals AC and DB which meet at point E.
  • So, point E being the common mid-point of both diagonals.

As both of the diagonals of square are equal in length.

And the length of BE is 5 units, which is half the length of BD. It means the diagonal BD has a length of 10 units.

And since both diagonals  AC and DB are equal in length, it means the length of AC must be 10 units.

Since, AC has the length of 10 units, it means the length of AE must be half of the length of AC. Thus, the indicated measure of the length of AE is 5 units.

Therefore, the indicated measure of the length of AE = 5 units

Keywords: square, diagonals, length, line segment

Learn more about square diagonals from brainly.com/question/628263

#learnwithBrainly

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3 years ago
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Lerok [7]

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Step-by-step explanation:

3 0
3 years ago
The sample space of a random experiment is {a, b, c, d, e} with probabilities 0.1, 0.1, 0.2, 0.4, and 0.2 respectively. Let A de
antoniya [11.8K]

Answer:

a) P(A) =P(a)+P(b) +P(c)= 0.1+0.1+0.2 = 0.4

b) P(B) =P(c) +P(d)+P(e)=0.2+0.4+0.2=0.8

c) P(A') = 1-P(A) =1-0.4=0.6

d) P(A \cup B) =0.4 +0.8-0.2 =1.0

e)  The intersection between the set A and B is the element c so then we have this:

P(A \cap B) = P(c) =0.2

Step-by-step explanation:

We have the following space provided:

S= [a,b,c,d,e]

With the following probabilities:

P(a) =0.1, P(b)=0.1, P(c) =0.2, P(d)=0.4, P(e)=0.2

And we define the following events:

A= [a,b,c], B=[c,d,e]

For this case we can find the individual probabilities for A and B like this:

P(A) = 0.1+0.1+0.2 = 0.4

P(B) =0.2+0.4+0.2=0.8

Determine:

a. P(A)

P(A) =P(a)+P(b) +P(c)= 0.1+0.1+0.2 = 0.4

b. P(B)

P(B) =P(c) +P(d)+P(e)=0.2+0.4+0.2=0.8

c. P(A’)

From definition of complement we have this:

P(A') = 1-P(A) =1-0.4=0.6

d. P(AUB)

Using the total law of probability we got:

P(A \cup B) =P(A) +P(B)-P(A \cap B)

For this case P(A \cap B) = P(c) =0.2, so if we replace we got:

P(A \cup B) =0.4 +0.8-0.2 =1.0

e. P(AnB)

The intersection between the set A and B is the element c so then we have this:

P(A \cap B) = P(c) =0.2

8 0
3 years ago
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