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Juliette [100K]
3 years ago
5

Which parent function is represented by the graph?

Mathematics
1 answer:
anzhelika [568]3 years ago
8 0

Answer:

absolute value parent function

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I need help pleeeeeeeeeeeeeease
dsp73

Answer:

Tj hit 85% of the balls pitched to him

Step-by-step explanation:

6 0
3 years ago
Evaluate 3^2 + (8 - 2) x 4 - 6/3
labwork [276]

3^2 + (8 - 2) x 4 - 6/3

=9 + 24 - 2

=33 - 2

=31

4 0
3 years ago
Read 2 more answers
The total cost for paperback books varies
postnew [5]

Answer:

$113.81

Step-by-step explanation:

Find the cost for one book:

35.94 / 6 = 5.99

Multiply 5.99 with the number of books buying:

5.99 x 19 = 113.81

4 0
2 years ago
Required information Skip to question NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to re
pogonyaev

The result for the given 14 length bit string is-

(a) The bit strings of length 14 which have exactly three 0s is 2184.

(b) The bit strings of length 14 which have same number of 0s as 1s is 17297280.

(c) The bit strings of length 14 which have at least three 1s is 4472832.

<h3>What is a bit strings?</h3>

A bit-string would be a binary digit sequence (bits). The size of the value is the amount of bits contained in the sequence.

A null string is a bit-string with no length.

The concept used here is permutation;

ⁿPₓ = n!/(n-x)!

Where, n is the total samples.

x is the selected samples.

(a) Because there are exactly three 0s, its other digits have always been one, hence the total of permutations is equal to;

n = 14 and x = 3 digits.

¹⁴P₃ = 14!/(14-3)!

¹⁴P₃ = 2184.

Thus, the bit strings of length 14 which have exactly three 0s is 2184.

(b) Because it is a bit string with 14 digits and it can only have digits 0 or 1, we must choose 7 0s and the remaining 7 1s, hence the number possible permutations is equal to;

n = 14 and x = 7 digits.

¹⁴P₇ = 14!/(14-7)!

¹⁴P₇ = 17297280.

Thus, the bit strings of length 14 which have same number of 0s as 1s is 17297280.

(c) The answer is identical to problem a, but the rest of a digits could be either 0 or 1, hence it must be doubled by 2¹¹ because there are 11 digits, each of which can be one of two options;

= 2184×2¹¹

= 4472832

Thus, he bit strings of length 14 which have at least three 1s is 4472832.

To know more about permutation, here

brainly.com/question/12468032

#SPJ4

The correct question is-

How many bit strings of length 14 have:

(a) Exactly three 0s?

(b) The same number of 0s as 1s?

(d) At least three 1s?

3 0
1 year ago
Question 2 - Signal and System Properties. State whether each of the statements is true or false. Note that a statement is true
gregori [183]

Answer:

  (a) true

  (b) true

  (c) false; {y = x, t < 1; y = 2x, t ≥ 1}

  (d) false; y = 200x for .005 < |x| < 1

Step-by-step explanation:

(a)  "s(t) is periodic with period T" means s(t) = s(t+nT) for any integer n. Since values of n may be of the form n = 2m for any integer m, then this also means ...

  s(t) = s(t +2mt) = s(t +m(2T)) . . . for any integer m

This equation matches the form of a function periodic with period 2T.

__

(b) A system being linear means the output for the sum of two inputs is the sum of the outputs from the separate inputs:

  s(a) +s(b) = s(a+b) . . . . definition of linear function

Then if a=b, you have

  2s(a) = s(2a)

__

(c) The output from a time-shifted input will only be the time-shifted output of the unshifted input if the system is time-invariant. The problem conditions here don't require that. A system can be "linear continuous time" and still be time-varying.

__

(d) A restriction on an input magnitude does not mean the same restriction applies to the output magnitude. The system may have gain, for example.

6 0
3 years ago
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