To make it easier, lets say Justin = J, Eva = E, and Emma = M. Then, J = E + 7.50 M = J -12 J + E + M = 63 You can substitute for M in the third equation using the second equation, and you get: J + E + J - 12 = 63 Clean it up a little bit and you get: 2J + E = 75 You know J = E + 7.50, so substitute that for J and you get: 2(E + 7.50) + E = 75 2E + 15 + E = 75 3E + 15 = 75 3E = 60 E = 20 ; meaning that Eva has $20. Justin has $7.50 more than Eva, so he has $27.50. Emma has $12 less than Justin, so Emma has $15.50. Double check your answer by making sure they all add up to 63, which they do.
Answer:
Owen can afford to keep and drive the car for 4 days.
Step-by-step explanation:
The total owing on the car rental is R(d) = ($77.25/day)d + ($0.12/mile)m ≤ $330. Substitute 1 for d (that is, the rental is for 1 day) and 175 for m:
R(d) = ($77.25/day)(d days) + ($0.12/mile)(175 miles) ≤ $330
= $77.25d + $21 ≤ $330
This simplifies to $77.25d + $21 ≤ $330, or
$77.25d ≤ $309
Solving this by dividing both sides of the above equation by $77.25, we get
d = ($309)/($77.25) = 4
Owen can afford to keep and drive the car for 4 days.
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![s(av) = \frac{∆x}{∆t} \\](https://tex.z-dn.net/?f=s%28av%29%20%3D%20%20%5Cfrac%7B%E2%88%86x%7D%7B%E2%88%86t%7D%20%20%5C%5C%20)
![s(av) = average \: \: speed](https://tex.z-dn.net/?f=s%28av%29%20%3D%20average%20%5C%3A%20%20%5C%3A%20speed)
![∆x = covered \: \: distance](https://tex.z-dn.net/?f=%E2%88%86x%20%3D%20covered%20%5C%3A%20%20%5C%3A%20distance)
![∆t = time \: \: of \: \: the \: \: covered \: \: distance \\](https://tex.z-dn.net/?f=%E2%88%86t%20%3D%20time%20%5C%3A%20%20%5C%3A%20of%20%5C%3A%20%20%5C%3A%20the%20%5C%3A%20%20%5C%3A%20covered%20%5C%3A%20%20%5C%3A%20distance%20%5C%5C%20)
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![s(av) = \frac{183}{3} \\](https://tex.z-dn.net/?f=s%28av%29%20%3D%20%20%5Cfrac%7B183%7D%7B3%7D%20%20%5C%5C%20)
![s(av) = 61 \: \: \frac{mile}{hour}](https://tex.z-dn.net/?f=s%28av%29%20%3D%2061%20%5C%3A%20%20%5C%3A%20%20%5Cfrac%7Bmile%7D%7Bhour%7D%20)
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Using the normal distribution, it is found that approximately 40% of the scores are greater than 413.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
In this problem, we have that the mean and the standard deviation of the scores are given by:
![\mu = 400, \sigma = 50](https://tex.z-dn.net/?f=%5Cmu%20%3D%20400%2C%20%5Csigma%20%3D%2050)
Approximately 40% of the scores are greater than the 60th percentile, which is <u>X when Z = 0.253</u>.
Then:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![0.253 = \frac{X - 400}{50}](https://tex.z-dn.net/?f=0.253%20%3D%20%5Cfrac%7BX%20-%20400%7D%7B50%7D)
X - 400 = 50(0.253)
X = 412.65.
Rounding up, approximately 40% of the scores are greater than 413.
More can be learned about the normal distribution at brainly.com/question/24663213
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