Which point is located at (2, -5)?

Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o

kirza4 [7]

Answer:

a) H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

b) $df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

c) $t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

$\sigma_o =1.34$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic (chi square test)

$\alpha=0.01$ significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

The statistic is given by:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

Part b

The degrees of freedom are given by:

$df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

Part c

Replacing the info we got:

$t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

5 0

2 years ago

Consider the following theorem and proof.Theorem: The number âš2 is not rational number.Proof: Let's suppose âš2 is a rational n

ololo11 [35]

Answer:

The statement "Both of the numbers a and b cannot be even." is justified by the fact that a/b is simplified lowest terms

Step-by-step explanation:

We need to show that the √2 is an irrational number.

And from the given steps of proof stated in the question, we need to find the assumption that justifies the fact : " Both of the number a and b cannot be even".

First take the given options :

Option a : √2 is a rational number

√2 being an rational or irrational has no relation of a and b to be even or odd. So, this option is rejected.

Option B : a/b is simplified lowest terms

This shows that a and b are not even because if a and b are even then a/b can be simplified in other lowest term.

Option c : √2 is a irrational number

Similarly, By using the inverse part of Option A, option c is also rejected.

Option d : The fact that b divides a evenly

This only shows that the a is even. This does not give any idea about b is even or not. So option D is also rejected.

Option E : The fact that a and b are whole numbers

This fact does not imply that the a and b are even or odd. So option E is also rejected.

Hence, The statement "Both of the numbers a and b cannot be even." is justified by the fact that a/b is simplified lowest terms

7 0

2 years ago

HELP PLEASE HELP HELP

kupik [55]

I think it’s 0but not s

5 0

1 year ago

Write a rule to describe the function shown. x y −6 −4 −3 −2 0 0 3 2 y equals start fraction two over three end fraction x y equ

sertanlavr [38]

Answer:

$y=\frac{2}{3} x$ (y equals start fraction two over three end fraction x)

Step-by-step explanation:

Lest's organize the table values of our function first:

x y

-6 -4

-3 -2

0 0

3 2

The simplest kind of of function is a linear function of the form $y=mx+b$ where $m$ is the slope and $b$ is the y-intercept. Let's us check if our function is a linear one finding its equation and checking that all points are in the line.

The first step to find a linear equation is find the slope of the line; to do it, we are using the slope formula:

$m=\frac{y_2-y_1}{x_2-x_1}$

where

$m$ is the slope of the line

$(x_1,y_1)$ are the coordinates of the first point

$(x_2,y_2)$ are the coordinates of the second point

We know from our table that the first and second points are (-6, -4) and (-3, -2) respectively, so $x_1=-6$ , $y_1=-4$ , $x_2=-3$ , and $y_2=-2$ .

Replacing values:

$m=\frac{y_2-y_1}{x_2-x_1}$

$m=\frac{-2-(-4)}{-3-(-6)}$

$m=\frac{-2+4}{-3+6)}$

$m=\frac{2}{3)}$

Now that we have the slope of our line we can use the point slope formula to complete our linear function:

$y-y_1=m(x-x_1)$

where

$m$ is the slope

$(x_1,y_1)$ are the coordinates of the first point

Replacing values:

$y-y_1=m(x-x_1)$

$y-(-4)=\frac{2}{3)}(x-(-6))$

$y+4=\frac{2}{3)}(x+6)$

$y+4=\frac{2}{3} x+4$

$y=\frac{2}{3} x+4-4$

$y=\frac{2}{3} x$

Now, to check if our function is valid, we can either check if each point satisfies the rule $y=\frac{2}{3} x$ , or we can graph it and check if each lies is in the graph.

Let's check both:

We already know that points (-6, -4) and (-3, -2) satisfy the rule (after all, those were the point we used to came out with the rule in the first place), so we just need to check the points (0, 0) and (3, 2)

- For (0, 0):

$y=\frac{2}{3} x$

$0=\frac{2}{3}(0)$

$0=0$

The point satisfies the rule.

- For (3, 2):

$y=\frac{2}{3} x$

$2=\frac{2}{3} (3)$

$2=2$

The point satisfies the rule.

Since all the points of our table satisfies the rule $y=\frac{2}{3} x$ , we can conclude that the rule describing the function shown is $y=\frac{2}{3} x$ .

4 0

2 years ago

Read 2 more answers

Writing the Equation of a Parallel Line

Softa [21]

2x + 5y = - 5 is the answer. I m very expert in maths . So ask me your doubt. I can help

3 0

2 years ago