Greatest to least thats the order they need to be in

A CD usually sells for $14.00. If the CD is 20% off, and sales tax is 7%, what is the total price of the CD, including tax?

tankabanditka [31]

Answer:

the total price of the CD including tax is $11.984

Step-by-step explanation:

The computation of the total price of the CD including tax is shown below:

The Selling price of the CD is $14

CD is 20% off

And, the sales tax rate is 7%

So, the total price of the CD is

= ($14 - $14 × 20%) × (1.07)

= $11.2 × 1.07

= $11.984

Hence, the total price of the CD including tax is $11.984

8 0

2 years ago

10 in<br> Given the figure. Find AC.

rjkz [21]

Answer:

if its an arc and center angle is given it could be double the size. IF it is a triangle measure you would use Pythagoras for triangle and trig for given angle, if two triangles are shown and they are scale of each other alternative measures given divide into each measure by the correct line and check this with the matching angles. when found as a division this is the ratio so then you just multiply to find the larger measure but divide to find the smaller measure. AC could also be a junction or vector, if its a type of vector then you just follow the arrows and count how many arrows fit the line pick a direction and ie) if its x2 a then you show a+a.

Step-by-step explanation:

5 0

3 years ago

How far did this person travel in Section C ( hours 6-7) Remember to include units

cestrela7 [59]

From C in 1 hour they traveled 30 miles

5 0

3 years ago

A photoconductor film is manufactured at a nominal thickness of 25 mils. The product engineer wishes to increase the mean speed

Natali [406]

Answer:

$t=\frac{1.17-1.04}{\sqrt{\frac{0.11^2}{8}+\frac{0.09^2}{8}}}}=2.587$

$df=n_{1}+n_{2}-2=8+8-2=14$

Since is a one sided test the p value would be:

$p_v =P(t_{(14)}>2.587)=0.0108$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, we have enough evidence to reject the null hypothesis on this case and the 25 mil film have a mean greater than the 20 mil film so then the claim is not appropiate

Step-by-step explanation:

Data given and notation

$\bar X_{1}=1.17$ represent the mean for the sample 1 (25 mil film)

$\bar X_{2}=1.04$ represent the mean for the sample 2 (20 mil film)

$s_{1}=0.11$ represent the sample standard deviation for the sample 1

$s_{2}=0.09$ represent the sample standard deviation for the sample 2

$n_{1}=8$ sample size selected for 1

$n_{2}=8$ sample size selected for 2

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if reducing the film thickness increases the mean speed of the film, the system of hypothesis would be:

Null hypothesis: $\mu_{1} \leq \mu_{2}$

Alternative hypothesis: $\mu_{1} > \mu_{2}$

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{1.17-1.04}{\sqrt{\frac{0.11^2}{8}+\frac{0.09^2}{8}}}}=2.587$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{1}+n_{2}-2=8+8-2=14$

Since is a one sided test the p value would be:

$p_v =P(t_{(14)}>2.587)=0.0108$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, we have enough evidence to reject the null hypothesis on this case and the 25 mil film have a mean greater than the 20 mil film so then the claim is not appropiate

4 0

3 years ago

Urgent, Please help!

yarga [219]

Answer:

<h2>min: 52</h2><h2>max: 78</h2><h2>median: 63</h2><h2>lower quartile median: 54</h2><h2>upper quartile median: 73</h2><h2>interquartile range: 19</h2>

Step-by-step explanation:

the temperatures ordered from least to greatest

52, 53, 54, 54, 57, 62, 64, 69, 73, 73, 77, 78

the min is 52

the max is 78

median:

52, 53, 54, 54, 57, 62, 64, 69, 73, 73, 77, 78

(62+64)/2=126/2=63

thus the median is 63

now we don't include the median in finding the upper and lower quartile medians:

lower quartile median:

52, 53, 54, 54, 57, 62

54 (same #s no need to add up and divide by 2)

upper quartile median:

64, 69, 73, 73, 77, 78

73 (same #s no need to add up and divide by 2)

interquartile range is:

upper quartile median-lower quartile median:

73-54

19

8 0

3 years ago