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2 years ago

If you know the answer to this question please help!! ASAP I will give brainiest! Fr

Mathematics

1 answer:

leonid [27]2 years ago

6 0

Answer:

In five years, the enrollment would have been 10,724 students

Step-by-step explanation:

The general depreciation formula is given as;

P = I( 1 - r)^t

where P is the present population

I is the initial

r is the rate of depreciation = 4.5% = 4.5/100 = 0.045

t is 5 years

substituting these values;

P = 13,500 (1- 0.045)^5

P = 10,723.84

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2 years ago

Which segment is a dilation of BC using A as the center of dilation and a scale factor of 2/3

BabaBlast [244]

Answer: 45

Step-by-step explanation:

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2 years ago

What is the LCM of 4,8,14?

Tamiku [17]

The LCM of 4,8,14 is 56

M of 4 are 4,8,12,16,20,24,28,32,36,40,44,48,52,56
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I hope this helps you

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3 years ago

PLEASE HELP ME. NO PHONY ANSWERS TY, ANSWER IF YOUR 100% RIGHT!

Ksivusya [100]

The relative frequency of female mathematics majors will be 0.5142.

<h3>How to find the relative frequency?</h3>

The proportion of the examined subgroup's value to the overall account is known as relative frequency.

A sample of 317 students at a university is surveyed.

The students are classified according to gender (“female” or “male”).

The table is given below.

Then the relative frequency of female mathematics majors will be

⇒ 36 / (36 + 34)

⇒ 36 / 70

⇒ 0.5142

Learn more about conditional relative frequency here:

brainly.com/question/8358304

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8 0

1 year ago

The city of Anvil is currently home to 21000 people, and the population has been growing at a continuous rate of 4% per year. Th

Yanka [14]

Answer:

They'll reach the same population in approximately 113.24 years.

Step-by-step explanation:

Since both population grows at an exponential rate, then their population over the years can be found as:

$\text{population}(t) = \text{population}(0)*(1 + \frac{r}{100})^t$

For the city of Anvil:

$\text{population anvil}(t) = 21000*(1.04)^t$

For the city of Brinker:

$\text{population brinker}(t) = 7000*(1.05)^t$

We need to find the value of "t" that satisfies:

$\text{population brinker}(t) = \text{population anvil}(t)\\21000*(1.04)^t = 7000*(1.05)^t\\ln[21000*(1.04)^t] = ln[7000*(1.05)^t]\\ln(21000) + t*ln(1.04) = ln(7000) + t*ln(1.05)\\9.952 + t*0.039 = 8.8536 + t*0.0487\\t*0.0487 - t*0.039 = 9.952 - 8.8536\\t*0.0097 = 1.0984\\t = \frac{1.0984}{0.0097}\\t = 113.24$

They'll reach the same population in approximately 113.24 years.

4 0

3 years ago