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Yuki888 [10]
2 years ago
12

A store is advertising a sale with 15% off all prices in the store. Sales tax is 8%. Which equation will correctly determine the

total cost, C, of buying an item with an original price of p, after the discount and sales tax are included? Select all that apply.
Mathematics
2 answers:
ICE Princess25 [194]2 years ago
4 0

Answer: 15%-8%=C

Step-by-step explanation:

docker41 [41]2 years ago
3 0

Answer:

Well, you didnt include answers, but... here's the explanation.

Step-by-step explanation:

Ok so, the sale is 15% off.  As an example, lets use $10 as the original price.  15% off of $10 = 1.5, so we subtract 10 - 1.5 = 8.5.  Now, if the sale tax is 8%, we need to add 8% of 10, which is 0.8 - Now we get our final equation to calculate p, which is 8.5 + 0.8 = P.  8.5 + 0.8 = 9.3, and so the final price after discount AND sales tax are included is $9.3 - Hope this helped.  Mark me as brainliest please :)

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A carnival has two payment options. Plan A, you pay $10 admission plus $3 for each ride. Plan B, you pay a $20 admission plus $1
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<span>Let x = # of rides

Plan A: 10 + 3x

Plan B: 20 + x

if x < 5 rides then plan A is better buy
if x = 5 both plans are the same
if x > 5 then plan B is the best buy

Prove:
x = 6 (rides)

plan A: </span>10 + 3x = 10 + 3(6) = 10+18 = $28
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2 years ago
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Which graph shows the solution to the system of linear inequalities? y &lt; 2x – 5 y &gt; –3x + 1
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Answer:

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Step-by-step explanation:


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2 years ago
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g An irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the
blagie [28]

Answer:

A 90% confidence interval of the true mean is [$119.86, $123.34].

Step-by-step explanation:

We are given that an irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the mean amount of money spent on textbooks was $121.60.

Also, the standard deviation of the population was $6.36.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                              P.Q.  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean amount of money spent on textbooks = $121.60

            \sigma = population standard deviation = $6.36

            n = sample of students = 36

            \mu = population mean

<em>Here for constructing a 90% confidence interval we have used One-sample z-test statistics as we know about population standard deviation.</em>

<em />

So, 95% confidence interval for the population mean, \mu is ;

P(-1.645 < N(0,1) < 1.645) = 0.90  {As the critical value of z at 5% level

                                                      of significance are -1.645 & 1.645}  

P(-1.645 < \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.645) = 0.90

P( -1.645 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < 1.645 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.90

P( \bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.90

<u>90% confidence interval for</u> \mu = [ \bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } } ]

                                      = [ 121.60-1.645 \times {\frac{6.36}{\sqrt{36} } } , 121.60+1.645 \times {\frac{6.36}{\sqrt{36} } } ]

                                      = [$119.86, $123.34]

Therefore, a 90% confidence interval of the true mean is [$119.86, $123.34].

5 0
2 years ago
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Answer:

  D. insufficient data

Step-by-step explanation:

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