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3 years ago

Helpppppppppppp plzzzzzzzzzzzzzzzzzzz

Mathematics

1 answer:

ZanzabumX [31]3 years ago

4 0

Answer:

Step-by-step explanation:

$\left[\begin{array}{ccc}7&31\\142&19\end{array}\right]$ = 7(19) - 142(31) = - 4269

$\left[\begin{array}{ccc}7&11&21\\55&-8&2\\-16&-1&-9\end{array}\right]$ = 7(- 8)(- 9) + 11(2)(- 16) + 21(55)(- 1) - 21(- 8)(- 16) - 7(2)(- 1) - 11(55)(- 9) = 1768

$\left[\begin{array}{ccc}9.07&6.02&2.01\\-30.7&2.5&3.5\\3.55&-1.1&2.35\end{array}\right]$ = 9.07(2.5)(2.35) + 6.02(3.5)(3.55) + 2.01(- 30.7)(- 1.1) - 2.01(2.5)(3.55) - 9.07(3.5)(- 1.1) - 6.02(- 30.7)(2.35) = 647.3561

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A sociologist was investigating the ages of grandparents of high school students. From a random sample of 10 high school student

umka2103 [35]

Answer:

$(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908$

$(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308$

So then the confidence interval for the difference of means is given by:

$-8.908 \leq \mu_M -\mu_F \leq 5.308$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_{M}= 69.8$ represent the mean for the age of grandmother

$\bar X_{F}= 71.6$ represent the mean for the age of grandfather

$s_{M}= 8.38$ represent the sample deviation for the age of grandmother

$s_{F}= 6.65$ represent the sample deviation for the age of grandfather

$n_M = n_F= 10$

Solution to the problem

For this case the confidence interval is given by:

$(\bar X_{M} -\bar X_F) \pm t_{\alpha/2} \sqrt{\frac{s^2_M}{n_M} +\frac{s^2_F}{n_F}}$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n_M +n_F-2=10+10-2=18$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,18)".And we see that $t_{\alpha/2}=2.101$

And replacing we got:

$(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908$

$(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308$

So then the confidence interval for the difference of means is given by:

$-8.908 \leq \mu_M -\mu_F \leq 5.308$

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