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finlep [7]
3 years ago
7

The admission fee at an amusement park is 20$ for children and 35$ adults. On a certain, day 264 people entered the park, and th

e admission fees collected totaled $6780. How many children and how many adults were admitted?
Mathematics
1 answer:
shtirl [24]3 years ago
4 0

Answer:

100 Adults and 164 Children

Step-by-step explanation:

To solve this we will need to write two equations to solve for the two unknowns. Firstly, we know that tickets are 20 for children and 35 for adults, and that the total price was 6780. We can write an expression to represent this. 6780 is the total price, this also means it is the price of all the adults + the price of all the children. The total price of the adults tickets is the price of each ticket multiplied by the number of adults, so the total price of adult tickets can be represented by: 35A, 35 being the ticket price per adult, and A being the number of adults. Similarly the total children's price can be expressed as such: 20C, 20 being the price per child, and C being the number of children. Since we know that these two values total to 6780, we can now form an equation:

35A (total adult price) + 20C (total children price) = 6780

35A+20C=6780

Now we need another expression so we can substitute for one of the values. The other piece of information we are given, is that a total of 264 people entered the park. This means that the total number of adults + the total number of children = 264. Since we already have variables for the numbers of adults and children we now have our second equation:

A (number of adults) + C (number of children) = 264

A+C=264

Now we can use this second equation to substitute for one of the values in the first equation. Let's substitute for children:

If we rearrange the second expression we will get:

A+C-A=264-A

C=264-A

So now we substitute this expression into our first formula:

35A+20C=6780

C=264-A

35A+20(264-A)=6780

Now since we only have one unknown, we can simplify, expand and solve!

35A+20*264-20*A=6780

35A+5280-20A=6780

35A-20A+5280-5280=6780-5280

15A=1500

15A/15=1500/15

A= 100

A total of 100 adults were admitted.

Now we can use the number of adults to find the number of children using our second formula:

C=264-A

C=264-100

C=164

And there we are! A total of 100 adults and 164 children were admitted.

Hope this helped!

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Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The price of these bulbs is very favorab
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Answer:

(a) Customer will not purchase the light bulbs at significance level of 0.05

(b) Customer will purchase the light bulbs at significance level of 0.01 .

Step-by-step explanation:

We are given that Light bulbs of a certain type are advertised as having an average lifetime of 750 hours. A random sample of 50 bulbs was selected,  and the following information obtained:

Average lifetime = 738.44 hours and a standard deviation of lifetimes = 38.2 hours.

Let Null hypothesis, H_0 : \mu = 750 {means that the true average lifetime is same as what is advertised}

Alternate Hypothesis, H_1 : \mu < 750 {means that the true average lifetime is smaller than what is advertised}

Now, the test statistics is given by;

       T.S. = \frac{Xbar-\mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, X bar = sample mean = 738.44 hours

               s  = sample standard deviation = 38.2 hours

               n = sample size = 50

So, test statistics = \frac{738.44-750}{\frac{38.2}{\sqrt{50} } } ~ t_4_9

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(a) Now, at 5% significance level, t table gives critical value of -1.6768 at 49 degree of freedom. Since our test statistics is less than the critical value of t, so which means our test statistics will lie in the rejection region and we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is smaller than what is advertised and so consumer will not purchase the light bulbs.

(b) Now, at 1% significance level, t table gives critical value of -2.405 at 49 degree of freedom. Since our test statistics is higher than the critical value of t, so which means our test statistics will not lie in the rejection region and we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is same as what it has been advertised and so consumer will purchase the light bulbs.

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