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Sedbober [7]
3 years ago
11

Mrs. Chin paid a 20% tip on the bill for lunch if the tip amount was $2.75 what was the bill for lunch before the tip was added

Mathematics
2 answers:
gulaghasi [49]3 years ago
8 0
13.75:))))))))))))))))))))
Lelu [443]3 years ago
5 0

Answer:

$13.75

Step-by-step explanation:

the bill before the tip is x.

0.20x = 2.75

now divide both sides by .20

x= 2.75/0.20

x=13.75

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<u>Supposing 60 out of 100 scores are passing scores</u>, the 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

  • The lower limit is 0.5.
  • The upper limit is 0.7.

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of \alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the z-score that has a p-value of \frac{1+\alpha}{2}.

60 out of 100 scores are passing scores, hence n = 100, \pi = \frac{60}{100} = 0.6

95% confidence level

So \alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 - 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.5

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 + 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.7

The 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

  • The lower limit is 0.5.
  • The upper limit is 0.7.

A similar problem is given at brainly.com/question/16807970

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The accessories cost $3000
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