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Lostsunrise [7]

3 years ago

7

If a cross section of the paperweight is cut perpendicular to the base but does not pass through the top vertex, which shape des

cribes the cross section? rectangle triangle trapezoid hexagon

1 answer:

Montano1993 [528]3 years ago

4 0

Rectangle shape paper weight

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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

3 years ago

Ona took a random sample of 20 soccer teams across Europe. She tracked the average number of goals

anygoal [31]

The predicted number of wins for a team that averages 1.5 goals per game will be 17 according to the regression model.

From the table given, the linear regression model for the data can be expressed as :

y = 14.02x - 3.64 ; where ;
y = Predicted number of wins ;
x = average number of goals per match ;
slope = 14.02 ;
Intercept = - 3.64

<u>The Number of wins for a team which averages 1.5 goals per match can be calculated thus</u> :

Average goals per match, x = 1.5

Substitute x = 1.5 into the equation :

y = 14.02(1.5) - 3.64

y = 17.39

y = 17 (nearest whole number)

Learn more :brainly.com/question/18405415

8 0

2 years ago

Write the linear equation that gives the rule for this table.

Maksim231197 [3]

to get the equation of any straight line all we need is two points, well, let's just grab them from the table.

hmmm let's use (-49 , -39) and hmmm say (67 , 77)

$(\stackrel{x_1}{-49}~,~\stackrel{y_1}{-39})\qquad (\stackrel{x_2}{67}~,~\stackrel{y_2}{77}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{77}-\stackrel{y1}{(-39)}}}{\underset{run} {\underset{x_2}{67}-\underset{x_1}{(-49)}}}\implies \cfrac{77+39}{67+49}\implies \cfrac{116}{116}\implies 1$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-39)}=\stackrel{m}{1}(x-\stackrel{x_1}{(-49)}) \\\\\\ y+39=1(x+49)\implies y+39=x+49\implies y=x+10$

8 0

2 years ago

How can you find 50% of any numbers

blagie [28]

Answer:

Divide it by two

Step-by-step explanation:

Divide it by two

4 0

3 years ago

Read 2 more answers

Show that there is no solution for the radical equation 4w+4=-4.

Ad libitum [116K]

Answer:

there is a solution

w= -2

Step-by-step explanation:

4w+4=-4

-4 -4

4w=-8

/4 /4

w=-2

3 0

3 years ago