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Alexandra [31]
4 years ago
6

50PTS!! does anyone know how you turn in a section of your slides my teacher wants us to turn in only a portion of our google sl

ides because we work on them each week(she says it gets confusing) if you know how to do this please let me know and will mark brainliest :). ​
Computers and Technology
1 answer:
HACTEHA [7]4 years ago
5 0

Answer:

The best way to turn in one part of google slides, since you can't split the slide when turning assignments on classroom, is to copy and paste the part you want to turn in into another slide and submit that portion. I've done for many projects before and if you name the different slides (Ex: Assignment Portion One 5/6/20) so it won't get confusing with the multiple slides.

This is the best way I can think of, I hope I helped, and please correct me if there is a better way!

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Which of the following rules need to be followed when using variables?<br> Choose all that apply.
Masja [62]

Answer:

- All variable names must begin with a letter of the alphabet or an. underscore( _ ).

-After the first initial letter, variable names can also contain letters and numbers.

-Uppercase characters are distinct from lowercase characters.

-You cannot use a C++ keyword as a variable name.

7 0
3 years ago
Read 2 more answers
The sum of all the minterms of a boolean function of n variables is equal to 1.
Aleksandr-060686 [28]

The answers are as follows:

a) F(A, B, C) = A'B'C' + A'B'C + A'BC' + A'BC + AB'C' + AB'C + ABC' + ABC

= A'(B'C' + B'C + BC' + BC) + A((B'C' + B'C + BC' + BC)

= (A' + A)(B'C' + B'C + BC' + BC) = B'C' + B'C + BC' + BC

= B'(C' + C) + B(C' + C) = B' + B = 1


b) F(x1, x2, x3, ..., xn) = ∑mi has 2n/2 minterms with x1 and 2n/2 minterms

with x'1, which can be factored and removed as in (a). The remaining 2n1

product terms will have 2n-1/2 minterms with x2 and 2n-1/2 minterms

with x'2, which and be factored to remove x2 and x'2, continue this

process until the last term is left and xn + x'n = 1

4 0
4 years ago
Sequentially prenumbered forms are an example of a(n): a. Processing control. b. Data transmission control. c. Input control. d.
Stella [2.4K]

Answer:

Sequentially pre-numbered forms are an example of a(n):

c. Input control.

Explanation:

  • Such a type of control in which keep updating data on the basis of monitoring of data is known as Processing Control. Data matching is an example of processing control.
  • Data Transmission Control is such a control in which transmission of data    is done. Parity check is an example of data transmission control.
  • Input Control is such type of control in which user can perform different tasks like adding text. Sequentially pre-numberered forms and turn around documents are an example of an input control.
  • Examples of Data entry control include batch total and validity check.

7 0
3 years ago
What were the technological innovations that allowed for the rapid expansion of the railroads?
liraira [26]

Answer:

There are different phases of railroad expansion with the innovations in technology.

Explanation:

Few of the technological innovations are described below that leads in railroad expansion more rapid.

1. Centralized Traffic control (CTC) is introduced in 1960's that is used to control the traffic on railroads using different signal control.

2. In 1990's after computer technology involvement, railway ticket and reservation system is automate and being centralized. That makes the railroad expansion improve.

3. Bullet train technology has been introduced, that makes the railway trains more faster.

4. Electric trains has been introduced to use green energy and reduce the dependency on the fuel to make environment clean and green.

5 0
3 years ago
Fifty-three percent of U.S households have a personal computer. In a random sample of 250 households, what is the probability th
aleksley [76]

Answer:

The correct Answer is 0.0571

Explanation:

53% of U.S. households have a PCs.

So, P(Having personal computer) = p = 0.53

Sample size(n) = 250

np(1-p) = 250 * 0.53 * (1 - 0.53) = 62.275 > 10

So, we can just estimate binomial distribution to normal distribution

Mean of proportion(p) = 0.53

Standard error of proportion(SE) =  \sqrt{\frac{p(1-p)}{n} } = \sqrt{\frac{0.53(1-0.53)}{250} } = 0.0316

For x = 120, sample proportion(p) = \frac{x}{n} = \frac{120}{250} = 0.48

So, likelihood that fewer than 120 have a PC

= P(x < 120)

= P(  p^​  < 0.48 )

= P(z < \frac{0.48-0.53}{0.0316}​)      (z=\frac{p^-p}{SE}​)  

= P(z < -1.58)

= 0.0571      ( From normal table )

6 0
3 years ago
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