Four common types of distractions are

Will somebody help me???? ANYBODY?? PLZZZZZZZZZZZZ

romanna [79]

I've never seen that third syntax. If it's legit, it probably does the same as I and II, which are the same.

6 0

4 years ago

What is a functional simulation?

Anika [276]

D) a type of training that allows...

4 0

4 years ago

6. Consider a circle of diameter d cm.

blagie [28]

Circles

Area of Circle is $\pi d^{2} /4$

Total percentage increase in the area of the modified circle is 69%

Explanation:

(a) Given the diameter of the circle be d

Then the radius of the circle = r = d/2

and the area of the circle is given by:

Area(A) = $\pi r^{2}$

A = $\pi (d/2)^{2}$

A = $\pi d^{2} /4$

The area of circle is $\pi d^{2} /4$

(b)

The increase in percentage while changing the dimensions is given by the formula :

Total % increase = a% + b% - (a% * b%)/100

Where ,

a% is the percentage increase in side a

b% is the percentage increase in side b

Let the percentage increase in radius is : a%

Since a% = b%

So the formula is

Total increase % = a% + a% + (a% * a%)/100

Given here the diameter of the circle increases by 30% so a =30

Therefore, putting the value of a in the formula

Total increase % = 30% + 30% + (30% *30%)/100

Total increase % = 60% + 9%

Total increase % =69%

Therefore the total percentage increase in the area of the modified circle is 69%

8 0

3 years ago

Universal Containers is designing a new community using the Customer Community license type. They would like to have the users c

motikmotik

Answer:

The correct option is B. Set all Organization-Wide Default security to be "Private" for both internal and external users and use Sharing Rules to grant the desired access.

Explanation:

Note: This question is not complete as the options are not included. The complete question with the options is therefore provided before answering the question as follows:

Universal Containers is designing a new community using the Customer Community license type. They would like to have the users complete survey questions through the community interface and store the responses in a Custom Object that has a lookup to the account object. Any internal user who has access to the account should be able to see all survey responses. All Customer Community users should be able to see surveys filled in by other users for their company, but not surveys for other companies. What are the correct security settings to achieve this?

A. Set all Organization-Wide Default security to be "Public Read/Write" for internal users and "Private" for external users.

B. Set all Organization-Wide Default security to be "Private" for both internal and external users and use Sharing Rules to grant the desired access

C. Set all Organization-Wide Default settings to be "Public Read/Write" for both internal and external users.

D. Set the custom object to be master-detail to the Account and leave the Organization-Wide Default settings as their default values.

The explanation of the answer is now provided as follows:

Organization Wide Default security refers to settings that provides most restrictive settings that may be opened up by Role Hierarchy and Role hierarchy can be opened by Sharing rules. And the visibility of records at the record level is determined by all of these factors. Private, Public Read, Read/Write, and Read/Write & Transfer are the four permissions available in Sharing Rules.

The rules that are used to offer sharing access to users in public groups, roles, or territories are referred to as sharing rules. By introducing automatic exceptions to your org-wide sharing policies, sharing rules allow certain users more access.

By implication, the correct security settings to achieve the objective stated in the question is to set all Organization-Wide Default security to be "Private" for both internal and external users and use Sharing Rules to grant the desired access.

Therefore, the correct option is B. Set all Organization-Wide Default security to be "Private" for both internal and external users and use Sharing Rules to grant the desired access.

5 0

3 years ago

2. A host computer with an IP address of 172.16.29.155 using a subnet mask of 255.255.255.0 and a gateway of 172.16.29.1 pings a

ASHA 777 [7]

Answer:

From the host IP address and subnet mask, you can tell that both computers are on the same network. Host 172.16.29.155 will need to send an ARP request broadcast for 172.16.29.83 to get the MAC address and update its ARP table if the host 172.16.29.83 has not already done so.

Explanation:

<h2>ADDRESS RESOLUTION PROTOCOL (ARP)</h2>

Address Resolution Protocol (ARP) is a procedure for mapping a logical address known as the IP address to a physical address known as MAC (Media Access Control) address in a local area network. The job of the ARP is essentially to translate 32-bit addresses to 48-bit addresses and vice-versa when using IP Version 4 (IPv4) to communicate on the network.

Most of the computer programs/applications use logical address (IP address) to send/receive messages, however, the actual communication happens over the physical address (MAC address) this happens in the layer 2 of the OSI model. An ARP request is a broadcast, and an ARP response is a Unicast.

Scenarios when ARP is used:

Scenario-1: The sender is a host and wants to send a packet to another host on the same network.

· Use ARP to find another host’s physical address

Scenario -2: The sender is a host and wants to send a packet to another host on another network.

· Sender looks at its routing table.

· Find the IP address of the next-hop (router) for this destination.

· Use ARP to find the router’s physical address

Scenario -3: the sender is a router and received a datagram destined for a host on another network.

· Router checks its routing table.

· Find the IP address of the next router.

· Use ARP to find the next router’s physical address.

Scenario -4: The sender is a router that has received a datagram destined for a host in the same network.

· Use ARP to find this host’s physical address.

The situation in the question above falls under scenario-1. From the host IP address and subnet mask, you can tell that both computers are on the same network. Host 172.16.29.155 will need to send an ARP request broadcast for 172.16.29.83 so to get the MAC address and update its ARP table if the host 172.16.29.83 has not already done so.

7 0

4 years ago