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katen-ka-za [31]
3 years ago
6

At Bridgeview School,7/10 of the total students are driven to school by their parents. Another 1/6 of the total students at Brid

geview School ride the bus. The remaining students walk. What fraction of the students walk to school.
Mathematics
1 answer:
jarptica [38.1K]3 years ago
5 0

Answer: 2/15

Step-by-step explanation:

We are given the information that at Bridgeview School, 7/10 of the total students are driven to school by their parents and that 1/6 of the total students at Bridgeview School ride the bus.

The fraction of the students that walk to school would be:

= 1 - (7/10 + 1/6)

= 1 - (21/30 + 5/30)

= 1 - 26/30

= 4/30

= 2/15

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A ski lift is designed with a total load limit of 20,000 pounds. It claims a capacity of 100 persons. An expert in ski lifts thi
Yanka [14]

Answer:

0.5 = 50% probability that a random sample of 100 independent persons will cause an overload

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 normal distributions can be 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.

For the sum of n values of a distribution, the mean is \mu \times n and the standard deviation is \sigma\sqrt{n}

An expert in ski lifts thinks that the weights of individuals using the lift have expected weight of 200 pounds and standard deviation of 30 pounds. 100 individuals.

This means that \mu = 200*100 = 20000, \sigma = 30\sqrt{100} = 300

If the expert is right, what is the probability that a random sample of 100 independent persons will cause an overload

Total load of more than 20,000 pounds, which is 1 subtracted by the pvalue of Z when X = 20000. So

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

Z = \frac{20000 - 20000}{300}

Z = 0

Z = 0 has a pvalue of 0.5

1 - 0.5 = 0.5

0.5 = 50% probability that a random sample of 100 independent persons will cause an overload

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3 years ago
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3 years ago
What is the length of this rectangle?
solong [7]

Answer:

long

Step-by-step explanation:

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6 0
3 years ago
Read 2 more answers
A tourist first walked 17km with a speed of v km/h. Then he hiked 12 km uphill with the speed that was 2 km/hour less than his o
ExtremeBDS [4]

Answer: t=\frac{17}{v}+\frac{12}{v-2}


Step-by-step explanation:

Given: A tourist first walked 17km with a speed of v km/h.

Since Speed=\frac{distance}{time}

therefore, Time=\frac{distance}{speed}

Let t_1 be the time he walked with speed v.

then t_1=\frac{17}{v}

Also he hiked 12 km uphill with the speed that was 2 km/hour less than his original speed.

Let t_2 be the time he hiked 12 km,

Then t_2=\frac{12}{v-2}

The total time for the whole trip is given by:-

t=t_1+t_2=

Substitute the values of t_1 and t_2 in the equation, we get

t=\frac{17}{v}+\frac{12}{v-2}

4 0
3 years ago
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