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Vlada [557]
3 years ago
10

Determine the situation that best describes a normal distribution.

Mathematics
2 answers:
Margaret [11]3 years ago
8 0

we know that

normal distributions:

a function that represents the distribution of many random variables as a symmetrical bell-shaped graph

It is a very common continuous probability distribution

we know that

heights of students in any class is continuous

because it can be any real numbers

so, option-A..........Answer

elixir [45]3 years ago
3 0

Answer:

a.heights of students in a class

Step-by-step explanation:

A normal distribution has following properties:

Normal distributions are symmetric and it has a bell shape graph and and the mean, median, and mode are all equal.

The shirt prices in a sale can be of the same price. The student grades can also be almost same in an easy test and the cost of similar kind of houses will have similar rates.

So, only height of students in a class are not same and they are continuous in nature.

So. the answer is option a.heights of students in a class,

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If the typical balance on Lucy's credit card is $650 and the interest rate (APR) on her credit card is 18%, how much in interest
Mariana [72]

$9.75

<h3>Further explanation   </h3>

Given:  

  • The typical balance on Lucy's credit card is $650.
  • The interest rate (APR) on her credit card is 18%.

Question:  

How much in interest would you expect Lucy to be charged in a typical month?

The Process:  

This problem includes the type of determining simple interest.

\boxed{ \ I = P \times r \times t \ }  

where,

  • I = simple interest
  • P = principal (initial amount)  
  • r = annual interest rate
  • t = time (in years)

This time we will find out how much in interest we would expect to be charged in a typical month.

The data is as follows:

  • P = 650
  • r = 18% or \frac{18}{100} or 0.18  
  • t = \frac{1}{12} year  (one month)

Let us calculate how much in interest we would expect to be charged in a typical month.

\boxed{ \ = 650 \times \frac{18}{100} \times \frac{1}{12} \ }  

\boxed{ \ = 65 \times \frac{3}{10} \times \frac{1}{2} \ }  

\boxed{ \ = \frac{195}{10} \times \frac{1}{2} \ }  

\boxed{ \ = \frac{97.5}{10} \ }  

\boxed{ \ = \frac{975}{100} \ }  

Thus the amount of interest we would expect Lucy to be charged in a typical month is $ 9.75.

_ _ _ _ _ _ _ _ _ _

Notes  

We must be able to distinguish between simple and compound interest. Please learn about this in the link attached below.

<h3>Learn more   </h3>
  1. What interest (in dollars) does he receive annually?  brainly.com/question/8192969    
  2. How much will our account accrue to in 9.5 years brainly.com/question/4127036  
  3. What is the total amount owed at the end of the 4 years brainly.com/question/13675435
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3 years ago
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Answer:

2-12i

Step-by-step explanation:

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Brainliest if correct!
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Answer:

Option c (letter V)

Step-by-step explanation:

V

5 0
3 years ago
Solve by substitution: <br> -5x + y = -2<br> -3x + 6y = -12<br><br> please explain
fomenos

First we pick the equation. Let's say we pick first one. From it we express y.

-5x+y=-2\Longrightarrow y=-2+5x

Then we use this y and plug it in instead of y in the second equation.

-3x+6(-2+5x)=-12

Now just solve for x.

-3x-12+30x=-12 \\-3x=-30x \\\underline{x=10}

We plug this x in the first equation.

-5\cdot10+y=-2

And solve for y.

-50+y=-2 \\\underline{y=48}

So the solution to the equation is geometrically a point P which lies on the intersection of the two lines.

\boxed{P(10,48)}

Hope this helps.

r3t40

4 0
4 years ago
The total monthly profit for a firm is P(x)=6400x−18x^2− (1/3)x^3−40000 dollars, where x is the number of units sold. A maximum
wlad13 [49]

Answer:

Maximum profits are earned when x = 64 that is when 64 units are sold.

Maximum Profit = P(64) = 2,08,490.666667$

Step-by-step explanation:

We are given the following information:P(x) = 6400x - 18x^2 - \frac{x^3}{3} - 40000, where P(x) is the profit function.

We will use double derivative test to find maximum profit.

Differentiating P(x) with respect to x and equating to zero, we get,

\displaystyle\frac{d(P(x))}{dx} = 6400 - 36x - x^2

Equating it to zero we get,

x^2 + 36x - 6400 = 0

We use the quadratic formula to find the values of x:

x = \displaystyle\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}, where a, b and c are coefficients of x^2, x^1 , x^0 respectively.

Putting these value we get x = -100, 64

Now, again differentiating

\displaystyle\frac{d^2(P(x))}{dx^2} = -36 - 2x

At x = 64,  \displaystyle\frac{d^2(P(x))}{dx^2} < 0

Hence, maxima occurs at x = 64.

Therefore, maximum profits are earned when x = 64 that is when 64 units are sold.

Maximum Profit = P(64) = 2,08,490.666667$

6 0
3 years ago
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