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professor190 [17]
3 years ago
6

a television and dvd player cost a total of $1056. the cost of the television is three times the cost of the dvd player. find th

e cost of each item
Mathematics
1 answer:
Makovka662 [10]3 years ago
5 0

The cost of TV is $ 792 and cost of DVD player is $ 264

<u>Solution:</u>

Given, total cost of television and dvd player = 1056

The cost of the television is three times the cost of the dvd player.  

Now, let the cost of DVD player be "n", then the cost TV will be (1056 – n)

And, we know that, cost of TV is 3 times Cost of DVD

\text { cost of Television } =3 \times \text { cost of DVD }

\begin{array}{l}{1056-n=3 \times n} \\\\ {1056=3 n+n} \\\\ {4 n=1056} \\\\ {n=264}\end{array}

So, the cost of DVD player is $264, the cost of TV will be (1056 - 264) = $792

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Answer:

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Step-by-step explanation:

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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".  

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\hat p =0.11 represent the proportion estimated of subjects reported with dizziness

n = 268 represent the random sample selected

\hat q = 1-\hat p = 1-0.11= 0.89 represent the proportion of subjects No reported with dizziness

E= 0.04 = 4% represent the margin of error for the confidence interval

\alpha= 1-0.90 =0.1 and this value represent the significance level of the test or the probability of error type I

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