Answer: 4 .
_____________

n/2 = 5 - 3 ;
_______________

n/2 = 2 ;

_______________

n = 2*2 ;
_______________

n = 4 .
_______________

7 0

4 years ago

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Simplify 9 to the 2nd over 9 to the 7th.

Gekata [30.6K]

Use the exponent rules
x^a/x^b =1/x^b-a

9^2/9^7 = 1 / 9^7-2 = 1/9^5
1 over 9 to the 5th = 59049
so it would be 1/59049
1 0ver 59049

7 0

4 years ago

A bank is trying to determine which model of safe to install. The bank manager believes that each model is equally resistant to

alexandr402 [8]

Answer:

So on this case the 95% confidence interval would be given by (-1.152;5.152).

$-1.152 < \mu_{safe1}-\mu_{safe2}$

Should we conclude that the average time it takes experts to crack the safes does not differ by model at the 5% significance level?

A) Yes, the 95% confidence interval for ?D contains 0.

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".

Solution

Let put some notation

x=value for Safe 1 , y = value for Safe 2

x: 103, 90, 64, 120, 104, 92, 145, 106, 76

y: 101, 94, 58, 112, 103, 90, 140, 110, 74

The first step is calculate the difference $d_i=x_i-y_i$ and we obtain this:

d: 2, -4, 6, 8, 1, 2, 5, -4, 2

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{18}{9}=2$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =4.093$

The next step is calculate the degrees of freedom given by:

$df=n-1=9-1=8$

Now we need to calculate the critical value on the t distribution with 8 degrees of freedom. The value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , so we need a quantile that accumulates on each tail of the t distribution 0.025 of the area.

We can use the following excel code to find it:"=T.INV(0.025;8)" or "=T.INV(1-0.025;8)". And we got $t_{\alpha/2}=\pm 2.31$

The confidence interval for the mean is given by the following formula:

$\bar d \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

Now we have everything in order to replace into formula (1):

$2-2.31\frac{4.093}{\sqrt{9}}=-1.152$

$2+2.31\frac{4.093}{\sqrt{9}}=5.152$

So on this case the 95% confidence interval would be given by (-1.152;5.152).

$-1.152 < \mu_{safe1}-\mu_{safe2}$

Should we conclude that the average time it takes experts to crack the safes does not differ by model at the 5% significance level?

A) Yes, the 95% confidence interval for D contains 0.

5 0

3 years ago

Which table describes the behavior of the graph of f(x) = 2x³ - 26x - 24?

Anna71 [15]

After you have plotted the graph of the polynomial, the x-intercepts of the given polynomial will be:
(-3,0),(-1,0) and (4,0)
therefore the interval will be:
(-∞,-3)-below, (-3,-1)-above,(-1,4)-below,(4,∞)-above
The answer is ]

4 0

3 years ago

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