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Rzqust [24]
3 years ago
6

Hans rented a truck for one day. There was a base fee of $15.95, and there was an additional charge of 77 cents for each mile dr

iven. Hans had to pay
$207,68 when he returned the truck. For how many miles did he drive the truck?
Mathematics
2 answers:
Monica [59]3 years ago
7 0

Answer: 249 miles

Step-by-step explanation:

<em>First write a function that represents the amount paid for renting a truck:</em>

  • Set x as each mile driven.
  • Set y as the total amount paid.
  • $15.95 is the base fee paid no matter the mile, meaning the rent start at $15.95, not 0.

<em>Function: y = mx + b</em>

  • m = slope = amount paid for each mile driven = 77¢ = $0.77
  • b = y-intercept = amount paid when 0 miles driven = base fee = $15.95

<u>y = 0.77x + 15.95</u>

<em>He paid a total of $207.68, therefore y = 207.68:</em>

207.68 = 0.77x + 15.95

<em>Solve the equation for x:</em>

207.68 - 15.95 = 0.77x

191.73 = 0.77x

x = 249 miles driven

geniusboy [140]3 years ago
4 0
6.75 because I did this already
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SAT scores are normed so that, in any year, the mean of the verbal or math test should be 500 and the standard deviation 100. as
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a) P(X>625)=P(\frac{X-\mu}{\sigma}>\frac{625-\mu}{\sigma})=P(Z>\frac{625-500}{100})=P(Z>1.25)

P(Z>1.25)=1-P(Z

b) P(400

P(-1

P(-1

c) z=-0.842

And if we solve for a we got

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So the value of height that separates the bottom 20% of data from the top 80% is 415.8.  

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 SAT scores of a population, and for this case we know the distribution for X is given by:

X \sim N(500,100)  

Where \mu=500 and \sigma=100

We are interested on this probability

P(X>625)

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>625)=P(\frac{X-\mu}{\sigma}>\frac{625-\mu}{\sigma})=P(Z>\frac{625-500}{100})=P(Z>1.25)

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

P(Z>1.25)=1-P(Z

Part b

We are interested on this probability

P(400

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(400

And we can find this probability with this difference:

P(-1

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

P(-1

Part c

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

P(X>a)=0.8   (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 0.2 of the area on the left and 0.8 of the area on the right it's z=-0.842. On this case P(Z<-0.842)=0.2 and P(Z>-0.842)=0.8

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=-0.842

And if we solve for a we got

a=500 -0.842*100=415.8

So the value of height that separates the bottom 20% of data from the top 80% is 415.8.  

8 0
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