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soldier1979 [14.2K]
3 years ago
12

All 231 students in the math club went on a field trip. some students rode in vans which hold 7 students each and some students

rode in buses which hold 25 students each. how many of each type of vehicle did they use if there were 15 vehicles total?
Mathematics
1 answer:
NeTakaya3 years ago
5 0
They used 8 vans and 7 buses.
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If x = 4 tan(θ), find sec(θ) in terms of x
prisoha [69]

Answer:

\sec \theta =  \pm \frac{ \sqrt{ {x}^{2}  + 16} }{4}

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\because \: x = 4 \tan \theta \\   \therefore\frac{x}{4}  =  \tan \theta....(1) \\ \\  \because \:  { \sec}^{2}  \theta = 1 +  { \tan}^{2}   \theta \\ \therefore \:   \sec \theta =   \pm\sqrt{1 +  { \tan}^{2}   \theta }  \\  \therefore \:   \sec \theta =  \pm\sqrt{1 +  {  \bigg(   \frac{x}{4} \bigg)}^{2}    }  \\ \therefore \:   \sec \theta =  \pm\sqrt{1 +  {   \frac{x^{2}}{16} }    }  \\  \therefore \:   \sec \theta =  \pm\sqrt{{   \frac{16 + x^{2}}{16} }    }  \\ \therefore \:   \sec \theta =  \pm \frac{ \sqrt{ {x}^{2}  + 16} }{4}

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Natasha_Volkova [10]

Step-by-step explanation:

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The shape of the distribution of the time required to get an oil change at a 20 minute oil change facility. However, records ind
Katyanochek1 [597]

Answer:

For a mean oil change time of 20.51 minutes there would be a​ 10% chance of being at or​ below

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 21.3, \sigma = 3.9, n = 40, s = \frac{3.9}{\sqrt{40}} = 0.6166

Treating this as a random​ sample, at what mean​ oil-change time would there be a​ 10% chance of being at or​ below?

This is the 10th percentile, which is X when Z has a pvalue of 0.1. So it is X when Z = -1.28.

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

By the Central Limit Theorem

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

-1.28 = \frac{X - 21.3}{0.6166}

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X = 20.51

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