All categories

3 years ago

You just have to find x. please help been stuck for so long.

Mathematics

1 answer:

zzz [600]3 years ago

5 0

Answer:

x = 84°

Step-by-step explanation:

The inscribed angle x is half the measure of the intercepted arc, that is

x = 0.5 × 168° = 84°

You might be interested in

Yesterday your bus ride took 10 minutes. TOday it took 13. What is the percent of increase?

Reika [66]

3 percent i think just double check

3 0

3 years ago

Read 2 more answers

What is 3 multiply by 3/5

Alecsey [184]

3×5 is 15
.................

6 0

3 years ago

Read 2 more answers

Please answer and simplify -5/8 + (-8/5)

alexandr402 [8]

Answer:

-89/40 or simplified it is -2 9/40

6 0

3 years ago

Read 2 more answers

What is 7/10 - 1/4 (fraction)

mylen [45]

7/10 - 1/4 use common denominators
28/40 - 10/40 = 18/40 = 9/20

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

3 years ago