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Irina-Kira [14]
4 years ago
7

At a local high school, enrollment in the drama club increases by 5 percent each year. What is the common ratio in this situatio

n?
Mathematics
2 answers:
Dovator [93]4 years ago
7 0

Answer:

common ratio will be 1.05.

Step-by-step explanation:

At a local high school, enrollment in the drama club increases by 5%.

Let initial number of students in high school is 'a'.

so next year increase in students will be = 5% of a

                                                                    = \frac{5}{100} × a

                                                                    = 0.05a

and total number of the students will be  = ( a + 0.05a )

                                                                    = ( 1.05a )

So common ratio of the geometric square formed will be

=  \frac{1.05a}{a}  =  1.05

Therefore, common ratio will be 1.05.

Nitella [24]4 years ago
6 0
(new enrollment) = (old enrollment) +5%*(old enrollment)
.. = (old enrollment)*(1 +5%)
.. = (old enrollment)*1.05

The common ratio is 1.05.
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Samuel compared the monthly rental rates of two-bedroom apartments in Beverly and Lowell over the years. The results are shown i
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Only the monthly rental cost of apartments in Beverly is changing exponentially.

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4 0
4 years ago
What is greater 9,000 g or 8 kg
nikdorinn [45]
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5 0
3 years ago
The operating costs for each machine for one day have an unknown distribution with mean 1610 and standard deviation 136 dollars.
Lelechka [254]

Answer:

The standard deviation for the sample mean distribution is \sigma_{\bar{X}}=\frac{136}{\sqrt{45}}=\frac{136\sqrt{5}}{15} \approx 20.274

Step-by-step explanation:

The central limit theorem states that if you have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population then the distribution of the sample means will be approximately normally distributed.

For the random samples we take from the population, we can compute the standard deviation of the sample means:

\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}

From the information given

The standard deviation σ = 136 dollars

The sample n = 45

Thus,

\sigma_{\bar{X}}=\frac{136}{\sqrt{45}}=\frac{136\sqrt{5}}{15} \approx 20.274

The standard deviation for the sample mean distribution is \sigma_{\bar{X}}=\frac{136}{\sqrt{45}}=\frac{136\sqrt{5}}{15} \approx 20.274

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4 years ago
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