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2 years ago

1. Three friends pooled their money to purchase a new game system that costs $298. One person

Mathematics

1 answer:

iragen [17]2 years ago

5 0

Answer:

First person: $107

Second person: $98

Third person: $93

Step-by-step explanation:

Let be "f" the amount of money (in dollars) that the first person contributed to the purchase, "s" the amount of money (in dollars) that the second person contributed to the purchase and "t" the amount of money (in dollars) that the third person contributed to the purchase.

With the information given in the exercise, you can set up the following equations:

Equation 1 → $f+s+t=298$

Equation 2 → $s=f-9$

Equation 3 → $t=f-14$

Substitute the Equations 2 and 3 into the Equation 1 and then solve for "f":

$f+(f-9)+(f-14)=298\\\\3f-23=298\\\\f=\frac{321}{3}\\\\f=107$

Finally, substitute the value of "f" into the Equation 2 and then into the Equation 3, in order to find the values of "s" and "t".

Therefore, you get:

$s=107-9\\\\s=98\\\\\\t=107-14\\\\t=93$

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The average rate of change (m) is the ratio of the change in function value to the width of the interval:
m = (f(6) - f(2))/(6 - 2)

To compute this, we need to compute f(6) and f(2).
f(6) = (0.25*6 -0.5)*6 +3.5 = 9.5
f(2) = (0.25*2 - 0.5)*2 +3.5 = 3.5
Then the average rate of change is
m = (9.5 - 3.5)/(6 - 2) = 6/4 = 1.5

The average rate of change is 1.5 thousand owners per year.

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2 years ago

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Children under 10 years and older people over 65 years receive a discount on movie tickets. Let x represent the age of a person

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2 years ago

The inside diameter of a randomly selected piston ring is a random variable with mean value 13 cm and standard deviation 0.08 cm

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Answer:

a) P(12.99 ≤ X ≤ 13.01) = 0.3840

b) P(X ≥ 13.01) = 0.3075

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the cental 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:

$\mu = 13, \sigma = 0.08$

(a) Calculate P(12.99 ≤ X ≤ 13.01) when n = 16.

Here we have $n = 16, s = \frac{0.08}{\sqrt{16}} = 0.02$

This probability is the pvalue of Z when X = 13.01 subtracted by the pvalue of Z when X = 12.99.

X = 13.01

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

By the Central Limit Theorem

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

$Z = \frac{13.01 - 13}{0.02}$

$Z = 0.5$

$Z = 0.5$ has a pvalue of 0.6915

X = 12.99

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

$Z = \frac{12.99 - 13}{0.02}$

$Z = -0.5$

$Z = -0.5$ has a pvalue of 0.3075

0.6915 - 0.3075 = 0.3840

P(12.99 ≤ X ≤ 13.01) = 0.3840

(b) How likely is it that the sample mean diameter exceeds 13.01 when n = 25?

P(X ≥ 13.01) =

This is 1 subtracted by the pvalue of Z when X = 13.01. So

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

$Z = \frac{13.01 - 13}{0.02}$

$Z = 0.5$

$Z = 0.5$ has a pvalue of 0.6915

1 - 0.6915 = 0.3075

P(X ≥ 13.01) = 0.3075

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w = 15 Divide

l = 3 + w Remember step 2?

l = 3 + 15 Replace the variable using your value for w

l = 18 Add

And you're done! Always check your work. It helps to create a picture of a rectangle while you're doing these problems as well. As you get used to these problems more and more, you can show more or less work than I've shown, but try to stay true to what the teacher asks of you. Good luck!

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Since we know the sample is not big enough to use a z-distribution, we use student's t-distribution instead.

The formula to calculate the confidence interval is given by:

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