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4 years ago

Please help. Solve 5x + 7 > 17.

Mathematics

2 answers:

valentina_108 [34]4 years ago

5 0

Answer:

$\large\boxed{\{x\ |\ x>2\}}$

Step-by-step explanation:

$5x+7>17\qquad\text{subtract 7 from both sides}\\\\5x+7-7>17-7\\\\5x>10\qquad\text{divide both sides by 5}\\\\\dfrac{5x}{5}>\dfrac{10}{5}\\\\x>2$

natta225 [31]4 years ago

3 0

This question is just like algebra. The gt sign is similar to an equal sign.
You want to isolate the variable (x) on one side.
And the numbers on the other side of the sign.
Then continue with algebra-
Make sure you don’t touch the sign or switch it.
You ONLY switch the sign if your dividing by a negative number.
Hope this helped.

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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

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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2 years ago