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3 years ago

Dev has 9 shells. Zoe has 55 shells. Zoe gives some shells to Dev. Now Zoe has 3 times as many as Dev.

2 answers:

8 0

Answer:let the initial number of shells that dev has be A, and initial number of shells that zoe has be B. let the number of shells that zoe gives to dev be x. after giiving x shells zoe is left with 3 times the number of shell as that of dev. therefore number of shells with zoe = 3×number of shells with dev.

Step-by-step explanation:

PtichkaEL [24]3 years ago

4 0

Might not be correct but I try to explain in my best ways(: question not from safari! let the initial number of shells that dev has be A, and initial number of shells that zoe has be B. let the number of shells that zoe gives to dev be x. after giiving x shells zoe is left with 3 times the number of shell as that of dev. therefore number of shells with zoe = 3×number of shells with dev.

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Answer:

H 0 : μ = 3.1 H 1 : μ > 3.1

And this test is defined a a right tailed test since the symbol for the alternative hypothesis H1 is >.

$t = \frac{3.13-3.1}{\frac{0.03}{\sqrt{75}}}= 8.66$

$df = n-1=75-1 =74$

The p value is:

$p_v = P(t_{74} >8.66) =3.65x10^{-13}$

The significance level is: 0.1 or 10%

And since the p value is very low compared to the significance level we have enough evidence to:

Reject the null hypothesis

Step-by-step explanation:

For this case we want to test the claim that mean GPA of night students is larger than 3.1 at the .10 significance level. The claim needs to be on the alternative hypothesis so then we have the following system of hypothesis:

H 0 : μ = 3.1 H 1 : μ > 3.1

And this test is defined a a right tailed test since the symbol for the alternative hypothesis H1 is >.

We have the following info given:

$\bar X = 3.13 , s =0.03 , n =75$

The statistic to check the hypothesis is given by:

$t = \frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}$

Replacing the info given we got:

$t = \frac{3.13-3.1}{\frac{0.03}{\sqrt{75}}}= 8.66$

The degrees of freedom are given by:

$df = n-1=75-1 =74$

The p value since is a right tailed ted is given by:

$p_v = P(t_{74} >8.66) =3.65x10^{-13}$

The significance level is 0.1 or 10%

And since the p value is very low compared to the significance level we have enough evidence to:

Reject the null hypothesis

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