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3 years ago

Nicole drove

Mathematics

2 answers:

levacccp [35]3 years ago

8 0

For this case, you want to know how many gallons of gasoline Nicole needs to drive 284 miles, knowing that with 17 she drove 355 miles. For this, we make a rule of three:

$355 \ miles -----------> 17 \ gallons\\284 \ miles -----------> x$

Where "x" represents the number of gallons consumed in 284 miles.

$x = \frac {284 * 17} {355}\\x = 13.6 \ gallons$

So, Nicole will need 13.6 gallons of gas to drive 284 miles

Answer:

13.6 gallons

zloy xaker [14]3 years ago

4 0

Answer:

13.6 i believe, but that could, should be right tho.

Step-by-step explanation:

355/17=20.88235294 then 284/20.88235294=13.6

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Probability that the mean service time is between 1 and 2 hours is 0.96764.

Step-by-step explanation:

We are given that a systematic random sample of 100 service appointments has been collected.

The 100 appointments showed an average preparation time of 90 minutes with a standard deviation of 140 minutes.

Let $\bar X$ = sample mean service time

The z-score probability distribution for sample mean is given by;

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where, $\mu$ = average preparation time = 90 minutes

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The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

Now, probability that the mean service time is between 60 and 120 minutes is given by = P(60 minutes < $\bar X$ < 120 minutes)

P(60 minutes < $\bar X$ < 120 minutes) = P( $\bar X$ < 120 min) - P( $\bar X$ $\leq$ 60 min)

P( $\bar X$ < 120 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{120-90}{\frac{140}{\sqrt{100} } }$ ) = P(Z < 2.14) = 0.98382

P( $\bar X$ $\leq$ 60 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{60-90}{\frac{140}{\sqrt{100} } }$ ) = P(Z $\leq$ -2.14) = 1 - P(Z < 2.14)