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
yanalaym [24]
2 years ago
7

What is the best estimate for 82% of 503?

Mathematics
1 answer:
AleksAgata [21]2 years ago
5 0
82% of 503 is 412.46 hope this helps
You might be interested in
An alarming number of U.S. adults are either overweight or obese. The distinction between overweight and obese is made on the ba
madreJ [45]

Answer:

(A) The probability that a randomly selected adult is either overweight or obese is 0.688.

(B) The probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C) The events "overweight" and "obese" exhaustive.

(D) The events "overweight" and "obese" mutually exclusive.

Step-by-step explanation:

Denote the events as follows:

<em>X</em> = a person is overweight

<em>Y</em> = a person is obese.

The information provided is:

A person is overweight if they have BMI 25 or more but below 30.

A person is obese if they have BMI 30 or more.

P (X) = 0.331

P (Y) = 0.357

(A)

The events of a person being overweight or obese cannot occur together.

Since if a person is overweight they have (25 ≤ BMI < 30) and if they are obese they have BMI ≥ 30.

So, P (X ∩ Y) = 0.

Compute the probability that a randomly selected adult is either overweight or obese as follows:

P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\=0.331+0.357-0\\=0.688

Thus, the probability that a randomly selected adult is either overweight or obese is 0.688.

(B)

Commute the probability that a randomly selected adult is neither overweight nor obese as follows:

P(X^{c}\cup Y^{c})=1-P(X\cup Y)\\=1-0.688\\=0.312

Thus, the probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C)

If two events cannot occur together, but they form a sample space when combined are known as exhaustive events.

For example, flip of coin. On a flip of a coin, the flip turns as either Heads or Tails but never both. But together the event of getting a Heads and Tails form a sample space of a single flip of a coin.

In this case also, together the event of a person being overweight or obese forms a sample space of people who are heavier in general.

Thus, the events "overweight" and "obese" exhaustive.

(D)

Mutually exclusive events are those events that cannot occur at the same time.

The events of a person being overweight and obese are mutually exclusive.

5 0
3 years ago
Find and interpret the mean absolute deviation of the data. $4.6,8.5,7.2,6.6,5.1,6.2,8.1,10.3$4.6,8.5,7.2,6.6,5.1,6.2,8.1,10.3​
trasher [3.6K]

Answer:

1.3845  

Step-by-step explanation:

i used this formula in the screenshot

and you can also try using calculator websites

3 0
3 years ago
Definitely confused..........
kirill [66]
Well how do you think i feel ?
3 0
3 years ago
A biased dice was rolled and the scores recorded in the table below
dusya [7]

Given:

The table of score and frequency.

To find:

The probability that on the next roll the score will be 4.

Solution:

From the given table it is clear that the score 4 occurs 10 times.

Number of times the dice was rolled = 17 + 24 + 23 + 10 + 16 + 30

                                                             = 120

The probability that on the next roll the score will be 4 is

\text{Required probability}=\dfrac{\text{Number of times 4 occurs}}{\text{Number of times the dice was rolled}}

\text{Required probability}=\dfrac{10}{120}

\text{Required probability}=\dfrac{1}{12}

Therefore, the required probability is \dfrac{1}{12}.

8 0
3 years ago
A psychology professor wants to know whether verbal ability is related to memory quality in current first-year students at her s
alexgriva [62]

Answer:

The  variable are

A and  B

Step-by-step explanation:

Generally we can define a variable as a name of a placeholder representing a value  now considering the option to be selected from we see that

    The students' verbal SAT scores is a variable  because it is a phrase or a place holder that represent a value (in the is case a  numerical value)  which the score its self

    The second option  The students' percentage of words that were correctly recognized on the original list is also a variable because it is a placeholder of a name (in this case a phrase ) that represented the actual value itself

  The  third option The 75 students is not a variable because it is not representing any value but itself is the value

  The  third option The 750 students is not a variable because it is not representing any value but itself is the value

8 0
3 years ago
Other questions:
  • 200m-100m+48,750=51,000-150m
    9·1 answer
  • Given that sin θ = x/4. Which expression represents θ in terms of x?
    6·1 answer
  • Combine like terms, need help ASAP
    7·1 answer
  • Please solve 1/3 ÷ 2/3​
    7·1 answer
  • Please help, our teacher wasn’t here and we got no explanation on how to do it
    13·2 answers
  • PLZZZ HELP!! BRAINLIEST FOR WHOEVER GETS RIGHT ANSWER
    5·1 answer
  • HELP PLEASE!!!!!!!!!!!!!!!
    8·2 answers
  • list the first three whole numbers greater than 50 that will have a remainder of 7 when you divide by 9
    15·1 answer
  • Multiply find the answer
    14·1 answer
  • Determine if true:<br> 3+2-1x4=15
    6·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!