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zzz [600]
3 years ago
8

Does anyone know how to execute this assignment on Scratch?

Computers and Technology
1 answer:
lapo4ka [179]3 years ago
5 0

Execute this assignment from Scratch in the following way

Explanation:

1.For each thread, first Scratch sets the 'active thread' to that thread. Then, it executes each block one by one in the stack for the active thread. It will execute the entire stack all in one go if it can.

2.The Hide block is a Looks block and a Stack block. If the block's sprite is shown, it will hide the sprite — if the sprite is already hidden, nothing happens. This block is one of the simplest and most commonly used Looks blocks.

3.Scratch is used in many different settings: schools, museums, libraries, community centers, and homes.

4.Mitch Resnik, the creator of the super-simple Scratch programming language and head of the Lifelong Kindergarten group at the MIT Media Lab, gave a TEDx talk about the value of coding and computer literacy in early education.

5.

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A factory producing cables for personal computers finds that its current average
puteri [66]
This is normal as personal computers, unless designed to, don't usually put out a ton of power it's mostly just what your battery need and a bit more that's it
8 0
3 years ago
[30 points, will mark Brainliest] Which of the following is the lowest hexadecimal value? Explain why. Options to chose; F2, 81,
larisa86 [58]

Answer:

The answer to this question is given below in the explanation section.

Explanation:

First, we need to convert these hexadecimal numbers into decimal numbers, then we can easily identify which one is the lowest hexadecimal.

The hexadecimal numbers are F2, 81, 3C, and 39.

F2 = (F2)₁₆ = (15 × 16¹) + (2 × 16⁰) = (242)₁₀

81 = (81)₁₆ = (8 × 16¹) + (1 × 16⁰) = (129)₁₀

3C = (3C)₁₆ = (3 × 16¹) + (12 × 16⁰) = (60)₁₀

39 = (39)₁₆ = (3 × 16¹) + (9 × 16⁰) = (57)₁₀

The 39 is the lowest hexadecimal number among the given numbers.

Because 39 hex is equal to 57 decimal.

39 = (39)₁₆ = (3 × 16¹) + (9 × 16⁰) = (57)₁₀

5 0
3 years ago
Assume that a large number of consecutive IP addresses are available starting at 198.16.0.0 and suppose that two organizations,
Fiesta28 [93]

Answer & Explanation:

An IP version 4 address is of the form w.x.y.z/s

where s = subnet mask

w = first 8 bit field, x = 2nd 8 bit field, y = 3rd 8 bit field, and z = 4th 8 bit field

each field has 256 decimal equivalent. that is

binary                                        denary or decimal

11111111      =        2⁸      =             256

w.x.y.z represents

in binary

11111111.11111111.11111111.11111111

in denary

255.255.255.255

note that 255 = 2⁸ - 1 = no of valid hosts/addresses

there are classes of addresses, that is

class A = w.0.0.0 example 10.0.0.0

class B = w.x.0.0 example 172.16.0.0

class C = w.x.y.0 example 198.16.8.1

where w, x, y, z could take numbers from 1 to 255

Now in the question

we were given the ip address : 198.16.0.0 (class B)

address of quantity 4000, 2000, 8000 is possible with a subnet mask of type

255.255.0.0 (denary) or

11111111.11111111.00000000.00000000(binary) where /s =  /16 That is no of 1s

In a VLSM (Variable Length Subnet Mask)

Step 1

we convert the number of host/addresses for company A to binary

4000 = 111110100000 = 12 bit

step 2 (subnet mask)

vary the fixed subnet mask to reserve zeros (0s) for the 12 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000            /16

variable subnet mask: 11111111.11111111.11110000.000000                /20

now we have added 4 1s in the 3rd field to reserve 12 0s

<u><em>subnet mask: 255.255.</em></u><u><em>16.</em></u><u><em>0 (where the 1s in each field represent a denary number as follows)</em></u>

<u><em>1st 1 = 128, 2nd 1 = 64 as follows</em></u>

<u><em>1        1       1      </em></u><u><em> 1 </em></u><u><em>      1     1     1     1</em></u>

<u><em>128  64     32    </em></u><u><em>16</em></u><u><em>    8    4     2    1</em></u>

step 3

in the ip network address: 198.16.0.0/19 <em>(subnet representation)</em> we increment this using 16

that is 16 is added to the 3rd field as follows

That means the ist Valid Ip address starts from

          Ist valid Ip add: 198.16.0.1 - 198.16.15.255(last valid IP address)

Company B starts<u><em>+16: 198.16.</em></u><u><em>16</em></u><u><em>.0 - 198.16.31.255</em></u>

<u><em>                   +16: 198.16.</em></u><u><em>32</em></u><u><em>.0- 198.16.47.255 et</em></u>c

we repeat the steps for other companies as follows

Company B

Step 1

we convert the number of host/addresses for company B to binary

2000 = 11111010000 = 11 bit

Step 2

vary the fixed subnet mask to reserve zeros (0s) for the 11 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000            /16

variable subnet mask: 11111111.11111111.11111000.000000                /21

now we have added 5 1s in the third field to reserve 11 0s

<u><em>subnet mask: 255.255.</em></u><u><em>8.</em></u><u><em>0 (where the 1s in each field represent a denary number as follows)</em></u>

<u><em>1st 1 = 128, 2nd 1 = 64 as follows</em></u>

<u><em>1        1       1       1       </em></u><u><em>1 </em></u><u><em>    1     1     1</em></u>

<u><em>128  64     32    16    </em></u><u><em>8 </em></u><u><em>   4     2    1</em></u>

Step 3

Starting from after the last valid Ip address for company A

in the ip network address: 198.16.16.0/21 (<em>subnet representation</em>) we increment this using 8

That means the ist Valid Ip address starts from

           Ist valid Ip add: 198.16.16.1 - 198.16.23.255(last valid IP address)

Company C starts <u><em>+16: 198.16.</em></u><u><em>24</em></u><u><em>.0- 198.16.31.255</em></u>

<em>                             </em><u><em> +16: 198.16.</em></u><u><em>32</em></u><u><em>.0- 198.16.112.255 et</em></u>c

Company C

Step 1

we convert the number of host/addresses for company C to binary

4000 = 111110100000 = 12 bit

Step 2

vary the fixed subnet mask to reserve zeros (0s) for the 12 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000            /16

variable subnet mask: 11111111.11111111.11110000.000000                /20

now we have added 4 1s in the 3rd field to reserve 12 0s

<u><em>subnet mask: 255.255.</em></u><u><em>16.</em></u><u><em>0 (where the 1s in each field represent a denary number as follows)</em></u>

<u><em>1st 1 = 128, 2nd 1 = 64 as follows</em></u>

<u><em>1        1       1       1       1     1     1     1</em></u>

<u><em>128  64     32    16    8    4     2    1</em></u>

Step 3

Starting from after the last valid ip address for company B

in the ip network address: 198.16.24.0/20 (subnet representation) we increment this using 16

That means the ist Valid Ip address starts from

           Ist valid Ip add: 198.16.24.1 - 198.16.39.255(last valid IP address)

Company C starts <u><em>+16: 198.16.40.0- 198.16.55.255</em></u>

<em>                          </em><u><em>    +16: 198.16.56.0- 198.16.71.255 et</em></u>c

Company D

Step 1

we convert the number of host/addresses for company D to binary

8000 = 1111101000000 = 13 bit

Step 2

vary the fixed subnet mask to reserve zeros (0s) for the 13 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000            /16

variable subnet mask: 11111111.11111111.11100000.000000                /19

now we have added 3 1s in the 3rd field to reserve 13 0s

<u><em>subnet mask: 255.255.</em></u><u><em>32.</em></u><u><em>0 (where the 1s in each field represent a denary number as follows)</em></u>

<u><em>1st 1 = 128, 2nd 1 = 64 as follows</em></u>

<u><em>1        1      </em></u><u><em> 1 </em></u><u><em>      1       1     1     1     1</em></u>

<u><em>128  64     </em></u><u><em>32  </em></u><u><em>  16    8    4     2    1</em></u>

Step 3

Starting from after the last valid ip address for company C

in the ip network address: 198.16.40.0/20 (subnet representation) we increment this using 32

That means the ist Valid Ip address starts from

           Ist valid Ip add: 198.16.40.1 - 198.16.71.255(last valid IP address)

Company C starts <u><em>+16: 198.16.72.0- 198.16.103.255</em></u>

<em>                          </em><u><em>    +16: 198.16.104.0- 198.16.136.255 et</em></u>c

5 0
3 years ago
What is the output of doublec= 12.0 / 5 Systemoutprintln (c)
artcher [175]

Answer:

The code will produce:-

2.4

Explanation:

In this code the result of the arithmetic operation is stored in the variable c.On evaluating the expression it divides 12.0 by 5 which results in 2.4 and it is stored in float variable c.Then it is printed on the screen using print statement.Since the c is double variable so the result will be a decimal number.

Hence the answer is 4.

6 0
3 years ago
If you posted an advertisement for your research study on marriage on the Internet and the couples who responded (i.e., the coup
WITCHER [35]

Answer:

B. volunteer bias.

Explanation:

-Experimenter bias is when the expectations of the experimenter in regards to the outcome are communicated to the participants in any form.

-Volunteer bias refers to a situation in which the people that volunteer to take place in a study doesn't represent the general public.

-Research bias is when the experimenter influence the result to get a specific outcome.

-Social desirability bias is when the people taking part in a study give their responses in a way that is viewed as favorable.

According to this, the answer is that the situation would be an example of volunteer bias.

3 0
3 years ago
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