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DedPeter [7]
2 years ago
8

What techniques overcome resistance and improve the credibility of a product? Check all that apply.

Computers and Technology
1 answer:
horrorfan [7]2 years ago
3 0

Answer: Including performance tests, polls, or awards.

Listing names of satisfied users

Explanation:

For every business, it is important to build ones credibility as this is vital on keeping ones customers and clients. A credible organization is trusted and respected.

The techniques that can be used to overcome resistance and improve the credibility of a product include having performance tests, polls, or awards and also listing the names of satisfied users.

Sending unwanted merchandise and also using a celebrity name without authorization is bad for one's business as it will have a negative effect on the business credibility.

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are designed to locate information based on the nature and meaning of Web content, not simple keyword matches Select one: a. Cra
san4es73 [151]

Answer: C. Semantic Search Engines

Explanation:

Semantic search is simply search with meaning. It is designed to locate information based on the nature and meaning of Web content, not simple keyword matches (like in lexical searches)

4 0
3 years ago
You are setting up your Windows computer to connect to the Internet and notice that when you type www.microsoft, the browser doe
Harrizon [31]

Answer:

Check your DNS settings

Explanation:

Different errors may occur when setting up your computer to connect to the internet, one of which is described in the question above.

The DNS (Domain Name System) is responsible for redirecting domain names to their physical IP address. Instead of remembering every IP address of sites you visit frequently, domain names are used for easy remembrance, the DNS makes the matching of domain names to IP addresses possible.

To change your DNS setting follow these steps:

  1. Click settings from your start menu
  2. Click on Network and Internet
  3. Look to the bottom of the main page and click on "Network and Sharing Center"
  4. On the left tab, click "Change adapter settings"
  5. Right Click on the current network you are using and select properties
  6. Left-click on the "Internet Protocol Version 4 (TCP/IPv4) and click on properties.
  7. Check to see if "Obtain DNS server address automatically" is selected, if it is selected,
  8. Click on the radio button under it "Use the following DNS server address"
  9. Enter the DNS address you want to use
  10. Click Ok and close the window.

The problem should be resolved.

4 0
3 years ago
What is the name of the mvost powerful battery
sergij07 [2.7K]
Optima battery because it is stronger than a factory battery
6 0
3 years ago
a cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits such that the giv
Leokris [45]

Using the knowledge in computational language in C++ it is possible to write a code that  cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits

<h3>Writting the code:</h3>

<em>#include <bits/stdc++.h></em>

<em>using namespace std;</em>

<em>// chracter to digit mapping, and the inverse</em>

<em>// (if you want better performance: use array instead of unordered_map)</em>

<em>unordered_map<char, int> c2i;</em>

<em>unordered_map<int, char> i2c;</em>

<em>int ans = 0;</em>

<em>// limit: length of result</em>

<em>int limit = 0;</em>

<em>// digit: index of digit in a word, widx: index of a word in word list, sum: summation of all word[digit]  </em>

<em>bool helper(vector<string>& words, string& result, int digit, int widx, int sum) { </em>

<em>    if (digit == limit) {</em>

<em>        ans += (sum == 0);</em>

<em>        return sum == 0;</em>

<em>    }</em>

<em>    // if summation at digit position complete, validate it with result[digit].</em>

<em>    if (widx == words.size()) {</em>

<em>        if (c2i.count(result[digit]) == 0 && i2c.count(sum%10) == 0) {</em>

<em>            if (sum%10 == 0 && digit+1 == limit) // Avoid leading zero in result</em>

<em>                return false;</em>

<em>            c2i[result[digit]] = sum % 10;</em>

<em>            i2c[sum%10] = result[digit];</em>

<em>            bool tmp = helper(words, result, digit+1, 0, sum/10);</em>

<em>            c2i.erase(result[digit]);</em>

<em>            i2c.erase(sum%10);</em>

<em>            ans += tmp;</em>

<em>            return tmp;</em>

<em>        } else if (c2i.count(result[digit]) && c2i[result[digit]] == sum % 10){</em>

<em>            if (digit + 1 == limit && 0 == c2i[result[digit]]) {</em>

<em>                return false;</em>

<em>            }</em>

<em>            return helper(words, result, digit+1, 0, sum/10);</em>

<em>        } else {</em>

<em>            return false;</em>

<em>        }</em>

<em>    }</em>

<em>    // if word[widx] length less than digit, ignore and go to next word</em>

<em>    if (digit >= words[widx].length()) {</em>

<em>        return helper(words, result, digit, widx+1, sum);</em>

<em>    }</em>

<em>    // if word[widx][digit] already mapped to a value</em>

<em>    if (c2i.count(words[widx][digit])) {</em>

<em>        if (digit+1 == words[widx].length() && words[widx].length() > 1 && c2i[words[widx][digit]] == 0) </em>

<em>            return false;</em>

<em>        return helper(words, result, digit, widx+1, sum+c2i[words[widx][digit]]);</em>

<em>    }</em>

<em>    // if word[widx][digit] not mapped to a value yet</em>

<em>    for (int i = 0; i < 10; i++) {</em>

<em>        if (digit+1 == words[widx].length() && i == 0 && words[widx].length() > 1) continue;</em>

<em>        if (i2c.count(i)) continue;</em>

<em>        c2i[words[widx][digit]] = i;</em>

<em>        i2c[i] = words[widx][digit];</em>

<em>        bool tmp = helper(words, result, digit, widx+1, sum+i);</em>

<em>        c2i.erase(words[widx][digit]);</em>

<em>        i2c.erase(i);</em>

<em>    }</em>

<em>    return false;</em>

<em>}</em>

<em>void isSolvable(vector<string>& words, string result) {</em>

<em>    limit = result.length();</em>

<em>    for (auto &w: words) </em>

<em>        if (w.length() > limit) </em>

<em>            return;</em>

<em>    for (auto&w:words) </em>

<em>        reverse(w.begin(), w.end());</em>

<em>    reverse(result.begin(), result.end());</em>

<em>    int aa = helper(words, result, 0, 0, 0);</em>

<em>}</em>

<em />

<em>int main()</em>

<em>{</em>

<em>    ans = 0;</em>

<em>    vector<string> words={"GREEN" , "BLUE"} ;</em>

<em>    string result = "BLACK";</em>

<em>    isSolvable(words, result);</em>

<em>    cout << ans << "\n";</em>

<em>    return 0;</em>

<em>}</em>

See more about C++ code at brainly.com/question/19705654

#SPJ1

3 0
2 years ago
In Rizzati Corp, vice presidents in departments such as engineering, manufacturing, IT, and human resources report directly to t
Harrizon [31]

Answer: <em>Functional organizational structure.</em>

Explanation:

From the given case/scenario, we can state that Rizzati Corp has a functional organizational structure. Functional organizational structure is referred to as a structure which is used in order to organize employees. The employees are mostly grouped or organized based on their knowledge or specific skills. It tends to vertically structure each and every department or staff with specific roles from VP to sales and finance departments, to worker assigned to commodities or services.

7 0
3 years ago
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