She should use a condition-controlled loop. Because it uses a True/False and it will not continue until the task is done
Answer:
True
Explanation:
Solution
ZigBee uses unlicensed frequency bands but operate at slower speed or data rates.
ZigBee: This communication is particular designed for control and sensor networks on IEEE 802.15.4 requirement for wireless personal area networks (WPANs), and it is a outcome from Zigbee alliance.
This communication level defines physical and (MAC) which is refereed to as the Media Access Control layers to manage many devices at low-data rates.
Answer: Well, in transit means "in the process of being transported" according to Merriam Webster, so I would think that would mean that it is being shipped from China over to America (Or wherever you may live) currently. Hope this helped!
Explanation:
Answer:
Explanation:
public class Main
{
private static String val; //current val
public static class TextInput
{
public TextInput()
{
val= new String();
}
public void add(char c)
{
if(val.length()==0)
{
val=Character.toString(c);
}
else
{
val=val+c;
}
}
public String getvalue()
{
return val;
}
}
public static class NumericInput extends TextInput
{
Override
public void add(char c)
{
if(Character.isDigit(c))
{
//if character is numeric
if(val.length()==0)
{
val=Character.toString(c);
}
else
{
val=val+c;
}
}
}
}
public static void main(String[] args)
{
TextInput input = new NumericInput();
input.add('1');
input.add('a');
input.add('0');
System.out.println(input.getvalue());
}
}
The fraction of shipments that will be accepted is 0.1299.
<h3>
How to calculate the probability?</h3>
Probability of a defective DVD = 0.04
Using Binomial distribution,
Fraction of shipments accepted = Probability of zero defects in sample of 50 = P(X = 0)
= 50C0 * 0.040 * (1 - 0.04)⁵⁰
= 0.96⁵⁰
= 0.1299
Fraction of shipments accepted = Probability of zero or one defects in sample of 50 = P(X = 0) + P(X = 1)
= 50C0 * 0.040 * (1 - 0.04)50-0 + 50C1 * 0.041 * (1 - 0.04)50-1
= 0.9650 + 50 * 0.04 * 0.9649
= 0.4005
Learn more about probability on:
brainly.com/question/24756209
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