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Wewaii [24]
3 years ago
6

PLEASEEEE HELPP Enrollment in a dance studio has grown exponentially since the studio opened. A graph depicting this growth is s

hown. Determine the percentage rate of growth
Enrollments
Time in Years
O 0.25%
1.25%
2.5%
25%

Mathematics
2 answers:
marishachu [46]3 years ago
8 0

Answer:

  25%

Step-by-step explanation:

The first doubling, from 20 to 40 occurs in a little more than 3 years, so the multiplier each year is a little less than 2^(1/3) ≈ 1.26. That is, each year is a little less than 26% more than the previous year.

The graph also goes near the point (8, 120), so grows by a factor of 6 in 8 years. That suggests a multiplier of 6^(1/8) ≈ 1.251. Each year is about 25.1% more than the previous year.

Both of these multipliers represent yearly growth rates near 25%.

___

Using the "rule of 72", the product of doubling time and percentage growth is about 72. So, for a doubling time of 3 years, the percentage growth is predicted to be near 72/3 = 24 percent.

All of these estimates help you choose the correct answer: 25%.

FinnZ [79.3K]3 years ago
7 0

Answer:

 25%

Step-by-step explanation:

The first doubling, from 20 to 40 occurs in a little more than 3 years, so the multiplier each year is a little less than 2^(1/3) ≈ 1.26. That is, each year is a little less than 26% more than the previous year.

The graph also goes near the point (8, 120), so grows by a factor of 6 in 8 years. That suggests a multiplier of 6^(1/8) ≈ 1.251. Each year is about 25.1% more than the previous year.

Both of these multipliers represent yearly growth rates near 25%.

___

Using the "rule of 72", the product of doubling time and percentage growth is about 72. So, for a doubling time of 3 years, the percentage growth is predicted to be near 72/3 = 24 percent.

All of these estimates help you choose the correct answer: 25%

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\dfrac{ {3}^{2} +  {4}^{3}  }{2 + 3}  \\  =  \dfrac{9 + 64}{5}  \\  =  \dfrac{73}{5}  \\  = 14.6

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Hope this helps. - M
3 0
3 years ago
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A​ half-century ago, the mean height of women in a particular country in their 20s was 64.7 inches. Assume that the heights of​
Ainat [17]

Answer:

99.5% of all samples of 21 of​ today's women in their 20's have mean heights of at least 65.86 ​inches.

Step-by-step explanation:

We are given that a half-century ago, the mean height of women in a particular country in their 20's was 64.7 inches. Assume that the heights of​ today's women in their 20's are approximately normally distributed with a standard deviation of 2.07 inches.

Also, a samples of 21 of​ today's women in their 20's have been taken.

<u><em /></u>

<u><em>Let </em></u>\bar X<u><em> = sample mean heights</em></u>

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

                          Z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean height of women = 64.7 inches

            \sigma = standard deviation = 2.07 inches

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 sample of 21 of​ today's women in their 20's have mean heights of at least 65.86 ​inches is given by = P(\bar X \geq 65.86 inches)

  P(\bar X \geq 65.86 inches) = P( \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } } \geq \frac{65.86-64.7}{\frac{2.07}{\sqrt{21} } } ) = P(Z \geq -2.57) = P(Z \leq 2.57)

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<em>The above probability is calculated by looking at the value of x = 2.57 in the z table which has an area of 0.99492.</em>

<em />

Therefore, 99.5% of all samples of 21 of​ today's women in their 20's have mean heights of at least 65.86 ​inches.

3 0
3 years ago
When a number is added to 1/5 of itself, the result is 24. The equation that models this problem is n +1/5 n = 24. What is the v
Lady_Fox [76]

For this case we must find the value of n of the following equation:

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A.

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