1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
OLga [1]
3 years ago
14

True or false: randomization allows each member of the population to have an equal probability of being selected for the sample

population
Mathematics
1 answer:
tia_tia [17]3 years ago
4 0

Answer:

True

Step-by-step explanation:

Simple random sampling chooses at random members of the population. This allows all members an equal probability of being selected for the sample.

You might be interested in
Current rules for telephone area codes allow the use of digits​ 2-9 for the first​ digit, and​ 0-9 for the second and third​ dig
Anna35 [415]

Using the fundamental counting theorem, we have that:

  • 648 different area codes are possible with this rule.
  • There are 6,480,000,000 possible 10-digit phone numbers.
  • The amount of possible phone numbers is greater than 400,000,000, thus, there are enough possible phone numbers.

The fundamental counting principle states that if there are p ways to do a thing, and q ways to do another thing, and these two things are independent, there are ways to do both things.

For the area code:

  • 8 options for the first digit.
  • 9 options for the second and third.

Thus:

8 \times 9 \times 9 = 648

648 different area codes are possible with this rule.

For the number of 10-digit phone numbers:

  • 7 digits, each with 10 options.
  • 648 different area codes.

Then

648 \times 10^7 = 6,480,000,000


There are 6,480,000,000 possible 10-digit phone numbers.

The amount of possible phone numbers is greater than 400,000,000, thus, there are enough possible phone numbers.

A similar problem is given at brainly.com/question/24067651

5 0
2 years ago
Which polynomial function has a leading coefficient of 1 and roots 21 and 31 with multiplicity 1?
schepotkina [342]

Answer:

Which i think is the first one, there may just be a typing error.

Step-by-step explanation:

A polynomial of order n has the following format:

f(x) = a(x - x_{0})(x - x_{1})...(x - x_{n-1})

In which a is the leading coefficient, x_{0}, x_{1},..., x_{n-1} are the roots.

If a root appears m times, they are said to have multiplicity m.

Leading coefficient of 1 and roots 21 and 31 with multiplicity 1

f(x) = 1(x - 21)(x - 31)

So the correct answer is:

f(x) = 1(x - 21)(x - 31)

Which i think is the first one, there may just be a typing error.

8 0
3 years ago
In right triangle ABC, B = 51° and AB = 14. BC = _____
liq [111]
The final angle has an angle measure of 115 degrees.
4 0
3 years ago
What is 12 and 19 /20 as a decimal?
-BARSIC- [3]

12.95

Hope this helps! ;)

6 0
3 years ago
Read 2 more answers
Please help! serious answers only
vovangra [49]
I believe also 115 degrees bc they’re vertical angles
5 0
3 years ago
Read 2 more answers
Other questions:
  • If the edge length of a cube is 13 feet, what is the correct way to write the expression to represent the volume of the cube in
    8·1 answer
  • Show work if needed.please help. Only the Canadian one.
    6·1 answer
  • If a chicken farm sold 3,000 eggs last week hoe many individual eggs were sold
    14·2 answers
  • Solve the equation <img src="https://tex.z-dn.net/?f=%2812%20x%5E%7B5%7Dy%2B12%20x%5E%7B6%7Dy-3%20y%5E%7B5%7D%20-18x%20y%5E%7B5%
    12·1 answer
  • Use 3.14 for ok to estimate the area of a circle. The diameter is given. Round your answer to the nearest hundredth if necessary
    15·1 answer
  • Help me for 1-4???????
    11·2 answers
  • Help for Geometry #6 and 7
    9·1 answer
  • Rama is stringing beads whose lengths are each 0.19 inches. How can rounding be used to estimate the length of a row of 9 beads?
    12·1 answer
  • Please Help!! 100 POINTS!!!
    14·2 answers
  • A) 32 x 3 = 3
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!