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____ [38]
3 years ago
5

PLEASE HELP ME

Mathematics
2 answers:
Aleonysh [2.5K]3 years ago
8 0

Answer:

(11)^{3}=1331

Step-by-step explanation:

The given expression is

log_{11}(1331)=3

To find the exponential form, we need the following property

log_{a}M=b \implies a^{b}=M

So, if we apply this property, we have

log_{11}(1331)=3 \implies (11)^{3}=1331

Which is completely true.

Therefore, the exponential form of the given logarithm is

(11)^{3}=1331

umka21 [38]3 years ago
7 0

log 11 (1331) =3

11^3 = 1331

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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

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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

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