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3 years ago

7

Identify the distance between the points (9,7,3) and (5,3,2), and identify the midpoint of the segment for which these are the e

ndpoints. Round to the nearest tenth, if necessary.

1 answer:

Daniel [21]3 years ago

8 0

Answer:

<em>dggkydkyrmhxmfjiildtlitslursusappurshfufzppufpu7sritstisitsuprspursur74s</em>

Step-by-step explanation:

she will is has p7w407wsdddfduufuur5e

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Help me pleaseeeeeeeeee <br>

Vera_Pavlovna [14]

Answer:

math po ba yun

Step-by-step explanation:

o sience anong subject yan

3 0

3 years ago

Read 2 more answers

Simplify-3(4a-5b)-(a-b)

Anuta_ua [19.1K]

$-3(4a-5b)-(a-b)=\\
-12a+15b-a+b=\\
-13a+16b$

5 0

3 years ago

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Solve for k.<br> 6k - 8 = 4K + 15

Brut [27]

Hi,

Answer:

k = $\frac{23}{2}$

Step-by-step explanation:

Subtract 4k from both sides

6k - 4k = 2k

2k - 8 = 15

Add 8 on both sides (you want to get rid of the 8 in order to leave the k alone)

2k = 23

k = 23/2

Have a good day!

4 0

3 years ago

Read 2 more answers

The random variable x is the number of occurrences of an event over an interval of ten minutes. It can be assumed that the proba

Olegator [25]

Answer:

The probability that occurrence of 8 in ten minutes is 0.0771.

Step-by-step explanation:

Poisson Distribution:

A discrete random variable X having the enumerable set {0,1,2,.....} as the spectrum, is said to have Poisson distribution with parameter $\mu$ (>0), if the p.m.f is given by

$P(X=x)=\frac{e^{-mu}\mu^x}{x!}$ for x=0,1,2,...

= 0, elsewhere

The mean number of occurrences in ten minutes is 5.3.

Here $\mu$ = 5.3 and x= 8

$P(X=8)=\frac{e^{-5.3}(5.3)^8}{8!}$

=0.0771

The probability that occurrence of 8 in ten minutes is 0.0771.

3 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