All categories

4 years ago

1. Target area range is a survey of what lies about __________ seconds ahead of the vehicle. A. 5-10 B. 10-20 C. 20-30 D. 40-50

Computers and Technology

2 answers:

pshichka [43]4 years ago

4 0

Answer:

Target area range is a survey of what lies about 12-20 seconds ahead of the vehicle. Correct Answer is B. (10-20 second)

Explanation:

Target area range is the space between your vehicle and the target area. A target area is an area on the roadway where the target is located. A target can be anything such as a car, a traffic signal, a blackhead, etc.

Target area range is also a section of roadway where the target is located and the area to the left and right of the target area

Anastaziya [24]4 years ago

4 0

Answer:

The correct answer is option B- 10-20

Explanation:

Target area range is a survey of what lies about 10-20 seconds ahead of the vehicle.

The target area range in a survey is kept at about 12 seconds to 20 seconds ahead of the vehicle. It is important to fix it at the center of the pathway as well. The area between the vehicle and the target area is known as the target area range. It could be anything, a vehicle, signal, or stationary objects.

You might be interested in

How do I go to files in Brainly I need help

Lyrx [107]

Answer:

There should be a little icon at the bottom of your question box/answer box. It looks like a paper clip. click it and boom

Explanation:

4 0

3 years ago

Create the following SQL Server queries that access the Northwind database. Place them in a zip file and place the zip file in t

neonofarm [45]

Answer:

1. SELECT e.EmployeeID, e.FirstName, e.LastName, COUNT(*) as OrderByEmployee FROM Employees e

JOIN Orders o

ON e.EmployeeID = o.EmployeeID

WHERE YEAR(o.OrderDate) = '1996'

GROUP BY e.EmployeeID,e.FirstName,e.LastName

ORDER BY OrderByEmployee DESC

2. SELECT o.OrderID,c.CustomerID, o.OrderDate,SUM((od.Quantity)*

od.UnitPrice - od.Discount)) as TotalCost FROM Orders o

JOIN [Order Details] od

ON od.OrderID = o.OrderID

JOIN Customers c

ON o.CustomerID = c.CustomerID

WHERE (MONTH(OrderDate)= 7 OR MONTH(OrderDate) = 8) and

YEAR(OrderDate) = 1996

GROUP BY o.OrderID,c.CustomerID, o.OrderDate

ORDER BY TotalCost DESC

3. SELECT c.CustomerID, c.CompanyName, c.City, COUNT(o.OrderID) as TotalOrder, SUM(od.Quantity* od.UnitPrice) as TotalOrderAmount FROM Customers c

JOIN Orders o

ON o.CustomerID = c.CustomerID

JOIN [Order Details] od

ON od.OrderID = o.OrderID

WHERE c.Country = 'USA'

GROUP BY c.CustomerID, c.CompanyName, c.City

ORDER BY c.City, c.CustomerID

4. SELECT c.CustomerID, c.CompanyName, c.City, COUNT(o.OrderID) as TotalOrder, SUM(od.Quantity* od.UnitPrice) as TotalOrderAmount FROM Customers c

JOIN Orders o

ON o.CustomerID = c.CustomerID

JOIN [Order Details] od

ON od.OrderID = o.OrderID

WHERE c.Country = 'USA'

GROUP BY c.CustomerID, c.CompanyName, c.City

HAVING COUNT(o.OrderID) > 3

ORDER BY c.City, c.CustomerID

5. SELECT p.ProductID, p.ProductName, s.ContactName as SupplierName, MAX(o.OrderDate) as LastOrderDateOfProduct FROM Products p

JOIN Categories c

ON c.CategoryID = p.CategoryID

JOIN Suppliers s

ON s.SupplierID = p.SupplierID

JOIN [Order Details] od

ON od.ProductID = p.ProductID

JOIN Orders o

ON o.OrderID = od.OrderID

where c.CategoryName = 'Grains/Cereals'

GROUP BY p.ProductID, p.ProductName, s.ContactName

ORDER BY SupplierName, p.ProductName

6. SELECT p.ProductID, p.ProductName, Count(DISTINCT c.CustomerID ) as OrderByThisManyDistinctCustomer

FROM Products p

JOIN [Order Details] od

ON od.ProductID = p.ProductID

JOIN Orders o

ON o.OrderID = od.OrderID

JOIN Customers c

ON c.CustomerID = o.CustomerID

where YEAR(o.OrderDate) = 1996

GROUP BY p.ProductID, p.ProductName

Explanation:

The six query statements returns data from the SQL server Northwind database. Note that all the return data are grouped together, this is done with the 'GROUP BY' Sql clause.

8 0

3 years ago

Please help me answering this question

Vanyuwa [196]

Answer:

none of them

Explanation:

the resistor is connected to the + pole of the battery, no circuit shows that.

a 1kΩ resistor would be brown-black-red, not brown black black.

8 0

2 years ago

HELP PLSSSSS!!! I WILL MARK BRAINLIEST FOR THE FIRST AND CORRECT ANSWER!!!

dalvyx [7]

The answer you are looking for would be A. Hardware can be an external tool, where as software is an internal tool. Hope this helped!

6 0

3 years ago

Read 2 more answers

Write a program that prompts the user to enter a series of numbers between 0 and 10 asintegers. The user will enter all numbers

Anna71 [15]

Answer:

The program in Python is as follows:

import collections

from collections import Counter

from itertools import groupby

import statistics

str_input = input("Enter a series of numbers separated by space: ")

num_input = str_input.split(" ")

numList = []

for num in num_input:

if(num.lstrip('-').isdigit()):

if num[0] == "-" or int(num)>10:

print(num," is out of bound - rejecting")

else:

print(num," is valid - accepting")

numList.append(int(num))

else:

print(num," is not a number")

print("Largest number: ",max(numList))

print("Smallest number: ",min(numList))

print("Range: ",(max(numList) - min(numList)))

print("Mode: ",end = " ")

freqs = groupby(Counter(numList).most_common(), lambda x:x[1])

print([val for val,count in next(freqs)[1]])

count_freq = {}

count_freq = collections.Counter(numList)

for item in count_freq:

print(item, end = " ")

lent= int(count_freq[item])

for i in range(lent):

print("#",end=" ")

print()

Explanation:

See attachment for program source file where comments are used as explanation. (Lines that begin with # are comments)

The frequency polygon is printed using #'s to represent the frequency of each list element

7 0

3 years ago