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2 years ago

11

An arts academy requires there to be 6 teachers for every 108 students and 3 tutors for every 36 students. students How many stu

dents does the academy have per teacher? Per tutor? How many tutors does the academy need if it has 84 students?

1 answer:

Alika [10]2 years ago

7 0

18 students = 1 teacher

13 students = 1 tutor

The academy needs 7 tutors for 84 students.

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he original price of a tractor was $45,000. The price of the tractor decreases at a steady rate of 8.3% per year. What is the va

Debora [2.8K]

Answer:

<em>$29178.25</em>

Step-by-step explanation:

<em>x = p(1 - r)ⁿ</em>

<em>p = original price</em>

<em>r = 8.3% = 0.083</em>

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3 years ago

John predicts the basketball team will score 68 points. If the basketball team

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3 years ago

there are 36 roses and 27 carnations. anna is making flower arrangements using both flowers, if she made the maximum number of a

solniwko [45]

This is a GCF problem,
find the GCF of 36 and 27. To do this, list out the factors for each number.
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3 years ago

In a simple random sample of 300 boards from this shipment, 12 fall outside these specifications. Calculate the lower confidence

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Answer:

The 95% confidence interval for the percentage of all boards in this shipment that fall outside the specification is (1.8%, 6.2%).

Step-by-step explanation:

In a random sample of 300 boards the number of boards that fall outside the specification is 12.

Compute the sample proportion of boards that fall outside the specification in this sample as follows:

$\hat p =\frac{12}{300}=0.04$

The (1 - <em>α</em>)% confidence interval for population proportion <em>p</em> is:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The critical value of <em>z</em> for 95% confidence level is,

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use a <em>z</em>-table.

Compute the 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification as follows:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.04\pm1.96\sqrt{\frac{0.04(1-0.04)}{300}}\\=0.04\pm0.022\\=(0.018, 0.062)\\\approx(1.8\%, 6.2\%)$

Thus, the 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification is (1.8%, 6.2%).

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2 years ago