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serious [3.7K]
3 years ago
7

What is mBD ? A.112 B.168 C.126 D.154

Mathematics
2 answers:
defon3 years ago
7 0

its bok guys bbbbbb

Marrrta [24]3 years ago
3 0
I think the correct answer is B
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Alex went bowling. He managed to get 3 strikes after 15 attempts. If you were to
algol13

Answer:

Theoretical Probability because its what should happen but experimental is what happens

Step-by-step explanation:

5 0
3 years ago
Using the bijection rule to count binary strings with even parity.
AleksandrR [38]

Answer:

Lets denote c the concatenation of strings. For a binary string <em>a</em> in B9, we define the element f(a) in E10 this way:

  • f(a) = a c {1} if a has an odd number of 1's
  • f(a) = a c {0} if a has an even number of 1's

Step-by-step explanation:

To show that the function f defined above is a bijective function, we need to prove that f is well defined, injective and surjective.

f   is well defined:

To see this, we need to show that f sends elements fromo b9 to elements of E10. first note that f(a) has 1 more binary integer than a, thus, it has 10. if a has an even number of 1's, then f(a) also has an even number because a 0 was added. On the other hand, if a has an odd number of 1's, then f(a) has one more 1, as a consecuence it will have an even number of 1's. This shows that, independently of the case, f(a) is an element of E10. Thus, f is well defined.

f is injective (or one on one):

If a and b are 2 different binary strings, then f(a) and f(b) will also be different because the first 9 elements of f(a) form a and the first elements of f(b) form b, thus f(a) is different from f(b). This proves that f in injective.

f is surjective:

Let y be an element of E10, Let x be the first 9 elements of y, then f(x) = y:

  • If x has an even number of 1's, then the last digit of y has to be 0, and f(x) = x c {0} = y
  • If x has an odd number of 1's, then the last digit of y has to be a 1, otherwise it wont be an element of E10, and f(x) = x c {1} = y

This shows that f is well defined from B9 to E10, injective, and surjective, thus it is a bijection.

3 0
3 years ago
Find the value of a and YZ if Y is between X and Z. XY = 7a, YZ = 5a, XZ = 6a + 24 = YZ =​
Aleksandr-060686 [28]
  • x-y-z

\\ \sf\longmapsto XY+YZ=XZ

\\ \sf\longmapsto 7a+5a=6a+24

\\ \sf\longmapsto 12a=6a+24

\\ \sf\longmapsto 12a-6a=24

\\ \sf\longmapsto 6a=24

\\ \sf\longmapsto a=\dfrac{24}{6}

\\ \sf\longmapsto a=5

  • YZ=5a=5(5)=25
5 0
2 years ago
(1+tanx)/(sinx+cosx)=secx
jek_recluse [69]

\frac{1+\tan x}{\sin x+\cos x}=\sec x is proved

<h3><u>Solution:</u></h3>

Given that,

\frac{1+\tan x}{\sin x+\cos x}=\sec x  ------- (1)

First we will simplify the LHS and then compare it with RHS

\text { L. H.S }=\frac{1+\tan x}{\sin x+\cos x}  ------ (2)

\text {We know that } \tan x=\frac{\sin x}{\cos x}

Substitute this in eqn (2)

=\frac{1+\frac{\sin x}{\cos x}}{\sin x+\cos x}

On simplification we get,

=\frac{\frac{\sin x+\cos x}{\cos x}}{\sin x+\cos x}

=\frac{\sin x+\cos x}{\cos x} \times \frac{1}{\sin x+\cos x}

Cancelling the common terms (sinx + cosx)

=\frac{1}{c o s x}

We know secant is inverse of cosine

=\sec x=R . H . S

Thus L.H.S = R.H.S

\frac{1+\tan x}{\sin x+\cos x}=\sec x

Hence proved

5 0
3 years ago
32 divied by y is equal to the product of three and y minus four
VLD [36.1K]

Answer:

your answer is y = 8/3

Step-by-step explanation:

6 0
2 years ago
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