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Illusion [34]
3 years ago
15

Please answer asap...

Mathematics
1 answer:
notka56 [123]3 years ago
4 0
B is the answer to this problem
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45
timofeeve [1]

Answer:

what should we find here

Step-by-step explanation:

confused

conform

5 0
3 years ago
Why is the greatest common factor of two numbers sometimes one
Gala2k [10]
Because it may be the only number that is common to both numbers. Like the greatest common factor between 2 and 3 is 1.
6 0
4 years ago
The ratio of boys to girls is 5 to 9. If there are 120 more girls than boys, how many boys are there? Can you please show work o
poizon [28]
5*30 = 150
9*30 = 270
Since the ratio is 5:9 you have to multiply both numbers in the ratio by the same constant.
I just started multiplying by 10, then 20, then 30 until I got numbers that had a difference of 120.
There are 150 boys.
Setting up an equation
5x + 120 = 9x
Solve for x and then multiply the ratio by the x to get the same result.
4 0
3 years ago
Bit transmission errors between computers sometimes occur, where one computer sends a 0 but the other computer receives a 1 (or
lana66690 [7]

Answer:

(a) Probability that a triplet is decoded incorrectly by the receiving computer. = 0.010

(b)

(1 – p) = 0.010

(c)

E(x) = 25000 x 0.010

     = 259.2

Explanation has given below.

Step-by-step explanation:

Solution:

(a) Probability that a triplet is decoded.

2 out of three

P = 0.94, n = 3

m= no of correct bits

m   bit (3, 0.94)

At p(m≤1) = B (1; 3, 0.94)

 = 0.010

(b) Using your answer to part (a),

(1 – p) = 0.010

Error for 1 bit transmission error.

(c)  How does your answer to part (a) change if each bit is repeated five times (instead of three?

P( m ≤ 2 )

L = Bit (5, 0.94)

   = B (2; 5, 0.94)

   = 0.002

(d)  Imagine a 25 kilobit message (i.e., one requiring 25,000 bits to send). What is the expected number of errors if there is no bit repetition implemented? If each bit is repeated three times?

Given:

h = 25000

Bits are switched during transmission between two computers = 6% = 0.06

m = Bit (25000, 0.06)

E(m) = np

        = 25000 x 0.06

         = 1500

m = Bit (25000, 0.01)

E(m) = 25000 x 0.010

     = 259.2

3 0
3 years ago
What else would need to be congruent to show that JKL= MNO by AAS?
taurus [48]
Answer: Choice B) Angle L = Angle O

---------------------------------------------------
---------------------------------------------------

If we know that Angle L is congruent to Angle O, then we can use the AAS (angle angle side) congruence property. We have one pair of angles marked by the square marker (angle J and angle M). So they are congruent angles. We have a pair of congruent sides JK = MN = 3. So we're just missing a pair of angles. 

Note: The answer is NOT angle K = angle N because this would mean ASA would be used instead of AAS. The order of the letters is important as it establishes how the sides and angles relate. With ASA, the side is between the angles. With AAS, the side is not between the angles. 

3 0
3 years ago
Read 2 more answers
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