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3 years ago

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A drive-in theater costs $20 per car, plus an additional fee of $4 per person. The cost, C, of the movie when p people are in th

e car can be represented by this equation: C = 20 + 4p Solve the equation for p to show How many people were in the car when the movie costed C dollars.

1 answer:

ozzi3 years ago

8 0

Answer:

$P=\frac{C}{4} -5$

Step-by-step explanation:

Step one:

given

cost per drive= $20

additional fee per person= $4

let the number of persons be p

and the total cost be C

The total cost function is expressed as

$C=20+4p$

Step two:

Required:

<u><em>We want to make P subject of the cost function </em></u>

Take 20 to the other side we have

C-20=4p

divide through by 4

P=C-20/4

simplify

P=C/4-5

$P=\frac{C}{4} -5$

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A culture started with 5,000 bacteria. After 8 hours, it grew to 6,000 bacteria. Predict how many bacteria will be present after

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Answer:

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It is known that the life of a particular auto transmission follows a normal distribution with mean 72,000 miles and standard de

scoray [572]

Answer:

a) $P(X$

$P(z$

b) $P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

$P(z>-0.583)=1-P(Z$

c) $P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

d) $z=1.28$

And if we solve for a we got

$a=72000 +1.28*12000=87360$

So the value of height that separates the bottom 90% of data from the top 10% is 87360.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the life of a particular auto transmission of a population, and for this case we know the distribution for X is given by:

$X \sim N(72000,12000)$

Where $\mu=72000$ and $\sigma=12000$

We are interested on this probability

$P(X$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X$

And we can find this probability using excel or the normal standard table and we got:

$P(z$

Part b

$P(X>65000)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>-0.583)=1-P(Z$

Part c

$P(X>100000)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

Part d

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.1$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.28$

And if we solve for a we got

$a=72000 +1.28*12000=87360$

So the value of height that separates the bottom 90% of data from the top 10% is 87360.

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3 years ago

Brad says that if a second number is 125% of the first number than the second number must be 75% of the second number is he corr

elena-s [515]

Answer:

Incorrect.

Step-by-step explanation:

Let, the first number is x, and the second number is y. Given, the second number is 125% of the first number. So, the equation will be

y = 1.25 x

Or, y= 5/4x

Now, multiplying it with 4/5, we get, x= 4/5y

Or, x= 0.80 y

But, Brad is saying the first number must be 75% of the second number. But here we see that the first number is 80% of the second number. That's why Brad is incorrect.

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3 years ago