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Find the length of the line segment with end points of 19 and -39

umka2103 [35]

I think to solve for this problem we need to subtract the given amounts.
19 - (-39) = 19 + 39 = 58

6 0

3 years ago

The percent, X , of shrinkage o n drying for a certain type of plastic clay has an average shrinkage percentage :, where paramet

never [62]

Answer:

a) $t=\frac{18.4-17.5}{\frac{2.2}{\sqrt{45}}}=2.744$

The degrees of freedom are given by:

$df=n-1=45-1=44$

The critical value for this case is $t_{\alpha}=1.68$ since the calculated value is higher than the critical we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 18.4

b) $p_v =P(t_{(44)}>2.744)=0.0044$

Step-by-step explanation:

Information given

$\bar X=18.4$ represent the sample mean

$s=2.2$ represent the sample standard deviation

$n=45$ sample size

$\mu_o =17.5$ represent the value to verify

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

Part a

We want to test if the true mean is higher than 17.5, the system of hypothesis would be:

Null hypothesis: $\mu \leq 17.5$

Alternative hypothesis: $\mu > 17.5$

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

And replacing we got:

$t=\frac{18.4-17.5}{\frac{2.2}{\sqrt{45}}}=2.744$

The degrees of freedom are given by:

$df=n-1=45-1=44$

The critical value for this case is $t_{\alpha}=1.68$ since the calculated value is higher than the critical we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 18.4

Part b

The p value would be given by:

$p_v =P(t_{(44)}>2.744)=0.0044$

5 0

3 years ago

1 third of the weight is ten

liraira [26]

Answer:

30

Step-by-step explanation:

10 times 3 = 30

3 0