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BaLLatris [955]
4 years ago
8

A store purchased a DVD for $12.00 and sold it to a customer for 50% more than the purchase price. The customer was charged a 7%

tax when the DVD was sold. What was the customer’s total cost for the DVD?
$12.84

$18.42

$18.84

$19
Mathematics
2 answers:
WITCHER [35]4 years ago
5 0
The answer is $19 or 19.26
olga55 [171]4 years ago
4 0

Answer:

$19.26 or $19.

Step-by-step explanation:

Sold DVD for 50% more of the purchase price of $12, so:

12 + 50/100 * (12) = 12 + 6 = 18.

The customer will pay 7% tax on $18.

18 * (1.07) = $19.26

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Find the equivalent fractions and highlight common denominators
Zepler [3.9K]

Answer:

smh

Step-by-step explanation:

too ez is the answer is 12421418258475172583521838716523/12312313123651253612534

3 0
3 years ago
Involving the selection of frozen dinners. Assume that there are 15 frozen dinners: 7 pasta, 6 chicken, and 2 seafood dinners. T
zzz [600]

The probability that at least 2 of the dinners selected are pasta dinners will be 0.8181...

<u><em>Explanation</em></u>

Pasta dinners = 7 , Chicken dinners = 6 and Seafood dinners = 2  

The student selects 5 of the total 15 dinners. So, total possible ways for selecting 5 dinners =^1^5C_{5} =3003

For selecting at least 2 of them as pasta dinners, the student can select 2, 3, 4 and 5 pasta dinners from total 7 pasta dinners.

So, the possible ways for selecting 2 pasta dinners =^7C_{2}*^8C_{3} =1176

The possible ways for selecting 3 pasta dinners =^7C_{3}*^8C_{2} =980

The possible ways for selecting 4 pasta dinners =^7C_{4}*^8C_{1} =280

The possible ways for selecting 5 pasta dinners =^7C_{5}*^8C_{0} =21

Thus, the probability for selecting at least 2 pasta dinners =\frac{1176+980+280+21}{3003} =\frac{2457}{3003} =0.8181.....

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4 years ago
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A researcher works on a study that has a population standard deviation of 0.73a sample mean of 84.5 and a sample size of 60 .Wha
Firlakuza [10]

Answer:

ME = z_{\alpha/2} \frac{\sigma}{\sqrt{n}}

The critical value can be founded in the normal standard distribution table using the value of \alpha/2 =0.025 and we got z_{\alpha/2}=1.96. Replacing the info given we got:

ME = 1.96 \frac{0.73}{\sqrt{60}}= 0.185

Step-by-step explanation:

For this case we have the following info given:

\sigma = 0.73 represent the population deviation

\bar X = 84.5 represent the sample mean

n =60 represent the sample size

And we want to find the margin of error for a confidence level of 95%. So then the significance level would be \alpha=1-0.95 = 0.05 and \alpha/2 =0.025. The margin of error is given by:

ME = z_{\alpha/2} \frac{\sigma}{\sqrt{n}}

The critical value can be founded in the normal standard distribution table using the value of \alpha/2 =0.025 and we got z_{\alpha/2}=1.96. Replacing the info given we got:

ME = 1.96 \frac{0.73}{\sqrt{60}}= 0.185

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