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3 years ago

.please help me with this I will crown you brainlest

Mathematics

2 answers:

Brut [27]3 years ago

8 0

I think it is either A or C
but more sure about A

Marizza181 [45]3 years ago

4 0

You can prove it by the SAS postulate which should be the first option

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MATHS MIDDLE SCHOOL ANSWER ASAP PLEASE

Sveta_85 [38]

Answer:

A. 1

B. -1

C. 5

Step-by-step explanation:

A. Using the Formula, we get u sub 2 = 9-9+1, which is just 1

B. Using the Formula, we get u sub 3 = 1-3+1, which is equal to -1

C. Using the Formula, we get u sub 4 = 1-(-3)+1, which is just 1+3+1 = 5

6 0

2 years ago

(6+8)^2=6^2+8^2 statement true

ddd [48]

this is false

(6+8)^2= 196

6^2+8^2=100

They are not equal so this is a false statement

5 0

3 years ago

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

Jayden drew a picture of a firepleace and then measured it. What is the area of jaydens drawing

nalin [4]

Send a pic of the fireplace

3 0

3 years ago

Write a quadratic equations given the roots 5/2 and 9

sukhopar [10]

Answer:

(x-5/2)(x-9).

x^2-5/2x-5/2x+45/2.

Multiply the number by 2.

which is 2(x^2-5/2x-5/2x+45/2).

2x^2-5x-5x+45.

2x^2-10x+45.

7 0

3 years ago