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ZanzabumX [31]
4 years ago
6

Average movie prices in the United States are, in general, lower than in other countries. It would cost $77.79 to buy three tick

ets in Japan plus two tickets in Switzerland. Three tickets in Switzerland plus two tickets in Japan would cost $73.91. How much does an average movie ticket cost in each of these countries?
Mathematics
1 answer:
sukhopar [10]4 years ago
6 0

Answer:

{Japan: $25.93} {Switzerland: $22.05}

Step-by-step explanation:

It's simple to solve for Japan but for Switzerland you got to do some reverse math.

For Japan all you gotta do is divide the given total by three because you're finding how much <u>one</u> ticket costs and not three. so there you have $25.93

For Switzerland here comes the reverse math. you have to take your $25.93 times 2 because is says "plus two tickets in Japan" so you have $51.86

Now that you've figured out the total of two tickets in Japan, you can subtract that from the given total ($73.91) and the answer is $22.05

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For Khloe's lemonade recipe, 8 lemons are required to make 10 cups of lemonade. How many lemons for 30 cups of lemonade?
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Answer: 24 lemons

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8 lemons - 10 cups

16 lemons - 20 cups

24 lemons - 30 cups

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Jarell Tarver purchased a $2,000.00 bond at the quoted price of 75 1/2. The bond paid 2 interest at a rate of 6 percent. What is
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3 years ago
A marine biologist monitors the population of sunfish in a small lake. He records 800 sunfish in his first year, 600 sunfish in
Colt1911 [192]

Answer:

Let's use the variable y to represent the number of years passed since the first year.

The population on the first year (y = 0) was 800

The population in the second year (y = 1) was 600.

The ratio in which the population decreased can be calculated as:

R = 600/800 = 0.75

This means that 600 is the 75% of 800, this also means that between the first year and the second year, the population decreased by the 25%

Now let's look at the ratio between the second and third year (y = 2), the population the third year was 450

Now the ratio is:

R = 450/600 = 0.75

Same as before, then we already can see that the population will decrease by 25% each year.

The generic exponential decay equation is:

f(y) = A*(1  - r)^y

where:

A = initial population, in this case, is 800

y = our variable, in this case, represents the number of years

r = the amount that decreases per each unit of our variable, this must be written in decimal form. In this case, we know that the population decreases by 25% each year, and 25% written in decimal form is 0.25

Also, (1 - r) = R, where R is the ratio we found earlier.

To do this, we just divide 25% by 100% to get (25%/100% = 0.25)

Then our equation will be:

f(y) = 800*(1 - 0.25)^y

f(y) = 800*( 0.75)^y

1) We want to predict when the population will be 200.

to do this, we set:

f(y) = 200 = 800*( 0.75)^y

and solve it for y.

(200/800) = 0.75^y

(1/4) = 0.75^y

Now we can use the relationship:

Ln(a^x) = x*ln(a)

Then let's apply Ln( ) in both sides to get

ln(1/4) = y*ln(0.75)

ln(1/4)/ln(0.75) = y = 4.8 years.

This means that 4.8 years after the first year, the population will be around 200.

2) The population in the 25 th year (this is 24 years after the first one, so we take y = 24) is:

f(24) = 800*(0.75)^(24) = 0.80

Around this point, we will have no more sunfish in the lake.

7 0
3 years ago
The average length of a field goal in the National Football League is 38.4 yards, and the s. d. is 5.4 yards. Suppose a typical
Tasya [4]

Answer:

a) The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 0.8538.

b) 0.25% probability that his average kicks is less than 36 yards

c) 0.11% probability that his average kicks is more than 41 yards

d-a) The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 1.08

d-b) 1.32% probability that his average kicks is less than 36 yards

d-c) 0.80% probability that his average kicks is more than 41 yards

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 38.4, \sigma = 5.4, n = 40, s = \frac{5.4}{\sqrt{40}} = 0.8538

a. What is the distribution of the sample mean? Why?

The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 0.8538.

b. What is the probability that his average kicks is less than 36 yards?

This is the pvalue of Z when X = 36. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{36 - 38.4}{0.8538}

Z = -2.81

Z = -2.81 has a pvalue of 0.0025

0.25% probability that his average kicks is less than 36 yards

c. What is the probability that his average kicks is more than 41 yards?

This is 1 subtracted by the pvalue of Z when X = 41. So

Z = \frac{X - \mu}{s}

Z = \frac{41 - 38.4}{0.8538}

Z = 3.05

Z = 3.05 has a pvalue of 0.9989

1 - 0.9989 = 0.0011

0.11% probability that his average kicks is more than 41 yards

d. If the sample size is 25 in the above problem, what will be your answer to part (a) , (b)and (c)?

Now n = 25, s = \frac{5.4}{\sqrt{25}} = 1.08

So

a)

The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 1.08

b)

Z = \frac{X - \mu}{s}

Z = \frac{36 - 38.4}{1.08}

Z = -2.22

Z = -2.22 has a pvalue of 0.0132

1.32% probability that his average kicks is less than 36 yards

c)

Z = \frac{X - \mu}{s}

Z = \frac{41 - 38.4}{1.08}

Z = 2.41

Z = 2.41 has a pvalue of 0.9920

1 - 0.9920 = 0.0080

0.80% probability that his average kicks is more than 41 yards

4 0
3 years ago
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