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marissa [1.9K]
3 years ago
13

In a college parking lot, the number of ordinary cars is larger than the number of sport utility vehicles by 59.3%. The differen

ce between the number of cars and the number of SUVs is 16. Find the number of SUVs in the lot.
Mathematics
1 answer:
frutty [35]3 years ago
4 0

Answer:

27 SUVs

Step-by-step explanation:

Let number of ordinary cars be x and SUVs be y

We can write 2 equations and use substitution to solve for the number of SUVs.

<u>"The number of ordinary cars is larger than the number of sport utility vehicles by 59.3%"- </u>

This means that 1.593 times more is ordinary cars (x) than SUVs (y), so we can write:

x  = 1.593y

<u>"The difference between the number of cars and the number of SUVs is 16" - </u>

Since we know ordinary cars are "more", we can say x - y = 16

<em>We can now plug in 1.593 y into x of the 2nd equation and solve for y:</em>

<em>x - y = 16</em>

<em>1.593y - y = 16</em>

<em>0.593y = 16</em>

<em>y = 27 (rounded)</em>

<em />

<em>Hence, there are 27 SUVs</em>

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That would be y = -2x+9

This is because each time y goes down by 2, x goes up by 1. We have slope = rise/run = -2/1 = -2. This indicates that the height of the candle decreases by 2 inches per hour. The slope represents the rate of change.

The initial height of the candle is the y intercept b value. So we have m = -2 and b = 9 lead us from y = mx+b to y = -2x+9

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The mean annual cost of an automotive insurance policy is normally distributed with a mean of $1140 and standard deviation of $3
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Using the normal distribution, it is found that the probabilities are given as follows:

a) 0.8871 = 88.71%.

b) 0.0778 = 7.78%.

c) 0.8485 = 84.85%.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

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

  • 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.
  • By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation s = \frac{\sigma}{\sqrt{n}}.

The parameters in this problem are given as follows:

\mu = 1140, \sigma = 310, n = 16, s = \frac{310}{\sqrt{16}} = 77.5

Item a:

The probability is the <u>p-value of Z when X = 1250 subtracted by the p-value of Z when X = 1000</u>, hence:

X = 1250:

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

By the Central Limit Theorem

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

Z = \frac{1250 - 1140}{77.5}

Z = 1.42

Z = 1.42 has a p-value of 0.9222.

X = 1000:

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

Z = \frac{1000 - 1140}{77.5}

Z = -1.81

Z = -1.81 has a p-value of 0.0351.

0.9222 - 0.0351 = 0.8871 = 88.71% probability.

Item b:

The probability is <u>one subtracted by the p-value of Z when X = 1250</u>, hence:

1 - 0.9222 = 0.0778 = 7.78%.

Item c:

The probability is the <u>p-value of Z when X = 1220</u>, hence:

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

Z = \frac{1220 - 1140}{77.5}

Z = 1.03

Z = 1.03 has a p-value of 0.8485.

0.8485 = 84.85% probability.

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1

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