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daser333 [38]
3 years ago
13

What is the answer for this math problem

Mathematics
1 answer:
nydimaria [60]3 years ago
6 0
Letter C is the answer
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Question 3 This questions requires you to examine computational differences for methods used to calculate sample variation. You
Licemer1 [7]

Answer:

sigma formulas are executed by using for loop to sum all the values.

public class H4_Q3{

public static void main(String[] args)

{

//Part a

double[] values = {100000000.6, 99999999.8, 100000002.8, 99999998.5, 100000001.3};

int n = values.length;

//Part b;

double sum = 0;

double sample_average = 0;

double sample_variance = 0;

int i = 0;

for(i = 0; i < n; i++)

{

sum = sum + values[i];

}

sample_average = sum/n;

for(i = 0; i < n; i++)

{

sample_variance = sample_variance + (Math.sqrt(values[i]) - sample_average);

}

System.out.println("Sample variance (Part b formula): "+sample_variance);

//Part c

double sum_squared = 0;

double sum_values = sample_average;

for(i = 0; i < n; i++)

{

sum_squared = sum_squared + Math.sqrt(values[i]);

}

sum_squared = sum_squared/n;

sample_variance = (sum_squared - Math.sqrt(sum_values)) * (n/ (n - 1));

System.out.println("Sample variance (Part c formula): "+sample_variance);

//Part d

sample_variance = 0;

double[] M = new double[n];

double[] S = new double[n];

M[0] = values[0];

S[0] = 0;

i = 1;

while(i < n)

{

M[i] = M[i-1] + ((values[i] - M[i-1])/i+1);

S[i] = S[i-1] + (values[i] - M[i-1]) * (values[i] - M[i]);

i++;

}

System.out.println("Sample variance (Part d formula): "+S[n-1]/n);

}

}

8 0
3 years ago
Does this graph show a function? Explain how you know.
Andrew [12]

Answer: C

Because it fails the vertical line test.

Put a pencil vertical on the graph.

Drag it along, if there are 2 lines that touch the pencil, it fails the test.

If there's only one, it passes.

5 0
3 years ago
Read 2 more answers
I need help with this i need u to help ne wit the number line
kupik [55]
Word the problem so it helps you with the double number line like this:
Question:
If 1/6 mile of a fence is completed each hour how long will it be in 4 hours?

Then use the double line:

1/6-----miles(1/6*4)------X  Length  X=1/6*4
  I                                                    X= 1/6*4/1
  I                                                    X= 4/6 ≈2/3 miles 
 1-------hours--------------4  Time
Here is a video that might help with your homework a lot :) Good luck
https://www.youtube.com/watch?v=89VhAPhvNcA

4 0
3 years ago
Q÷ 6 = 4 plzzzz helppp I need answerss ASAP
dalvyx [7]

Answer:

q=24

Step-by-step explanation:

the opposite of division is multiplication so 4x6=24. then you check this by of course imputing your answer into the equation. 24 ÷ 6 =4

4 0
3 years ago
Anybody help me to solve this question. ​
Mumz [18]

Answer:

\dfrac{1}{(b-c)},\dfrac{1}{(c-a)} ,\dfrac{1}{(a-b)} are\ in\ AP

Step-by-step explanation:

Given that (b-c)^2, (c-a)^2 , (a-b)^2 are in AP

To prove: \dfrac{1}{(b-c)},\dfrac{1}{(c-a)} ,\dfrac{1}{(a-b)} are in AP

From given as we know if p , q, r are in AP then 2q= p+r.

2(c-a)^2= (b-c)^2+(a-b)^2\\\\\Rightarrow 2(c^2+a^2-2ac)=b^2+c^2-2bc+a^2+b^2-2ab\\\\\Rightarrow 2c^2+2a^2-4ac= 2b^2+c^2+a^2 -2bc-2ab\\\\\Rightarrow a^2+c^2-2b^2-4ac= -2bc-2ab\\\\\Rightarrow a^2-2b^2+c^2= 4ac-2bc-2ab

Now

\dfrac{1}{(b-c)},\dfrac{1}{(c-a)} ,\dfrac{1}{(a-b)}2\dfrac{1}{(c-a)} =\dfrac{1}{(b-c)}+\dfrac{1}{(a-b)}\\\\\Rightarrow \dfrac{2}{(c-a)}= \dfrac{a-b+b-c}{(b-c)(a-b)} \\\\\Rightarrow \dfrac{2}{(c-a)}= \dfrac{a-c}{(b-c)(a-b)} \\\\\Rightarrow2(b-c)(a-b) = (c-a)(a-c) \\\\\Rightarrow 2(ab-b^2-ac+bc)= -(a-c)^2\\\\\Rightarrow 2ab- 2b^2-2ac+2bc = -a^2-c^2+2ac\\\\\Rightarrow a^2-2b^2+c^2=4ac-2ab-2bc

Which is the result of AP

.

Hence proved

6 0
3 years ago
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