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3 years ago

Kaylen wants to download 9 songs on iTunes. Each song costs $1.29. How much money does she need?

Mathematics

2 answers:

Anna35 [415]3 years ago

7 0

Answer:

$11.61

Step-by-step explanation:

Songs to be downloaded=9

Cost of Each=$1.29

So to solve this you will multiply By the amount of song by the cost of each.

So it's going to be;

$1.29×9

Which has its product as;

$11.61

So Therefore Kaylen will need $11.61

Veronika [31]3 years ago

6 0

Answer:

$11.61

Step-by-step explanation:

1.29 x 9=$11.61 its just multiplicationg, have a great day!!

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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.

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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.