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Dmitriy789 [7]
3 years ago
9

A museum opened at 9:00 a.m. In the first hour, 350 people purchased admission tickets. In the second hour, 20% more people purc

hased admission tickets than in the first hour. Each admission ticket cost $17.50. What was the total amount of money paid for all the tickets purchased in the first two hours?
Mathematics
2 answers:
charle [14.2K]3 years ago
7 0

Answer:

$13,475

Step-by-step explanation:

20% more is 100+20 = 120% of the previous

120% of 350

120/100 × 350

420

Total tickets in 2 hours:

350 + 420

770

Money paid:

770 × 17.50

13475

Furkat [3]3 years ago
3 0

Answer:

13,475

Step-by-step explanation:

First hour

350 people at $17.50 per ticket

350* $17.50 =$6125

Second hour

We add 20% more

350+350*.20 =350+70 =420

There were 420 people at $17.50

420*$17.50 =$7350

Add the total for the 2 hours

6125+7350=13,475

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Answer:

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Step-by-step explanation:

Let c represent the number of children ($1.75 each) and a represent the number of adults ( $2.00 each).

We know that there were 340 people total, so c + a = 340. This implies that a = 340 - c

We also know that $1.75 c + $2.00 a = $609.25

By substituting a with 340 -c we have $1.75 c + $2.00 (340 -c) = $609.25

Use the distributive property to obtain $1.75 c + $680 - $2.00 c = $609.25

Subtract $680 from both sides and combine like terms to get - $0.25 c = -

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2 years ago
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we have

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Answer:

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b) The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

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Step-by-step explanation:

If np' and n(1-p') are higher than 5, a confidence interval for the proportion is calculated as:

p'-z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }\leq  p\leq p'+z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }

Where p' is the proportion of the sample, n is the size of the sample, p is the proportion of the population and z_{\alpha/2} is the z-value that let a probability of \alpha/2 on the right tail.

Then, a 98% confidence interval for the percentage of all males who identify themselves as the primary grocery shopper can be calculated replacing p' by 0.52, n by 2400, \alpha by 0.02 and z_{\alpha/2} by 2.33

Where p' and \alpha are calculated as:

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