There are as many palindromic 12-bit binary strings as there are permutations with repetition of 6-bit binary strings, which is equal to

.
There are also only two permutations with repetition of 6-bit binary strings, that when each of them is mirrored and the result is "glued" to the right side of the original string, will give a palindromic string without 10 as a substring.
Those strings are 000000 and 111111.
Therefore there are

palindromic 12-bit binary strings with 10 as substring.
Answer:
0.6856
Step-by-step explanation:
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Now; assuming X = no of complaints received in a week
Required:
To find P(77 < X < 120)
Using a Gaussian Normal Distribution (
108,
= 20)
Using Z scores:

As a result X = 77 for N(108,20) is approximately equal to to Z = -1.75 for N(0,1)
SO;

Here; X = 77 for a N(108,20) is same to Z = 0.6 for N(0,1)
Now, to determine:
P(-1.75 < Z < 0.6) = P(Z < 0.6) - P( Z < - 1.75)
From the standard normal Z-table:
P(-1.75 < Z < 0.6) = 0.7257 - 0.0401
P(-1.75 < Z < 0.6) = 0.6856
C
the angle under y is 48 and line n adds up to 180
Answer: x1 = 0; x2 = 5/3; x3 = 1/3
x1 + 2x2 - x3 = 3 => 2x1 + 4x2 - 2x3 = 6
2x1 + x2 - 2x3 = 1
=> 4x2 - x2 = 5
⇔ 3x2 = 5
⇔ x2 = 5/3
with x2 = 5/3, we have: x1 - x3 = 3 - 10/3 = -1/3
3x1 - 5x3 = 5 - 4.5/3 = -5/3
=> x1 = 0
x3 = 1/3
Step-by-step explanation: