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s2008m [1.1K]
3 years ago
10

RSM was founded in 1996. From 1996 to 1997, the number of students in RSM doubled. From 1997 to 1998 the number of students doub

led again.
By what percent did the number of students increase from 1997 to 1998?
Mathematics
2 answers:
kaheart [24]3 years ago
3 0
To find the increase from 1997 to 1998, you could substitute in numbers and then determine the percent increase based on those numbers.

Ex: 500 to 1000 (96-97)
      1000 to 2000 (97-98)
Percent increase is (2000-1000)/1000 or 100% increase.

Testing another set of numbers will result in the same thing.

Ex: 200 to 400 (96-97)
      400 to 800 (97-98)
Percent increase is (800-400)400 or 100% increase.

The percent increase is 100%.
Neko [114]3 years ago
3 0

Answer:

The number of students increase from 1997 to 1998 by 100 %.

Step-by-step explanation:

Let x be the number of students in the year 1996.

According to the question,

The number of students on 1997 = 2x

Also, the number of students on 1998 = 4x

Thus, the percentage increase in the number of students from 1997 to 1998

=\frac{\text{the number of students on 1998-the number of students on 1997}}{\text{the number of students on 1997}}\times 100

=\frac{4x-2x}{2x}\times 100

=\frac{2x}{2x}\times 100

=100\%

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Answer:

2n-3=0

2n=3

n=3/2

therefore answer is 3/2

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Louis wants to carpet the rectangular floor of his basement. The basement has a area of 768 square feet. The width of the baseme
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Area=width*length. Let width=x ft, then length=3x ft, given the information in the problem. The area of basement=x*3x=768, 3x^2=768, x^2=256, x=16. The width is 16 feet. Therefore, length=3x=3*16=48 feet.
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The manager of a local nightclub has recently surveyed a random sample of 280 customers of the club. She would now like to deter
Lunna [17]

Answer:

z = \frac{35.6-35}{\frac{5}{\sqrt{280}}}= 2.007

p_v = P(z>2.007) = 0.0224

Since the p value is lower than the significance level given of 0.05 we have enough evidence to reject the null hypothesis on this case. And the best conclusion for this case is:

We (reject) the null hypothesis. That means that we (found) evidence to support the alternative.

Step-by-step explanation:

We have the following info given:

\bar X = 35.6 represent the sampel mean for the age of customers

\sigma = 5 represent the population standard deviation

n = 280 represent the sample size selected

We want to test if the mean age of her customers is over 35 so then the system of hypothesis for this case are:

Null hypothesis: \mu \leq 35

Alternative hypothesis \mu >35

The statistic for this case is given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing the data given we got:

z = \frac{35.6-35}{\frac{5}{\sqrt{280}}}= 2.007

We can calculate the p value since we are conducting a right tailed test like this:

p_v = P(z>2.007) = 0.0224

Since the p value is lower than the significance level given of 0.05 we have enough evidence to reject the null hypothesis on this case. And the best conclusion for this case is:

We (reject) the null hypothesis. That means that we (found) evidence to support the alternative.

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An engineer commutes daily from her suburban home to her midtown office. The average time for a one-way trip is 36 minutes, with
Ivanshal [37]

Answer:

57.93% probability that a trip will take at least 35 minutes.

Step-by-step explanation:

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.

In this problem, we have that:

\mu = 36, \sigma = 4.9

What is the probability that a trip will take at least 35 minutes

This probability is 1 subtracted by the pvalue of Z when X = 35. So

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

Z = \frac{35 - 36}{4.9}

Z = -0.2

Z = -0.2 has a pvalue of 0.4207

1 - 0.4207 = 0.5793

57.93% probability that a trip will take at least 35 minutes.

3 0
3 years ago
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