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4 years ago

A town doubles its size every 22 years. If the population is currently 4,500, what will the population be in 88 years?

1 answer:

5 0

So, the population will always be 50 percent of the previous population counted.
4,500×1.50^x=y
4,500 is the population that it starts with, while 1.50 is telling that it is increasing by 50%. X is the number of years.
We are trying to find it in 88 years, so we should replace x with 4. Here is the reason, it doubles every 22 years and we are looking 88 years.
4,500×1.50^4= 22,781.25
There can't be .25 of a person, so the answer would be 22,781.

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I think it’s B.33%

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3 years ago

At the beginning of a class period, half of the students in a class go to the library. Later in the period, half of the remainin

mash [69]

<u>Answer:</u>

32 students

<u>Step-by-step explanation:</u>

We are given that at the beginning of a class period, half of the students in a class go to the library and half of the remaining to the computer lab.

Given that there are 8 students remaining, we are to find the total number of students in the class initially.

At beginning = $x$ students

After half of them leave = $\frac{x}{2}$ students

After half of the remaining leave = $\frac{x}{4}$ students

So, $\frac{x}{4} = 8$

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Therefore, there were 32 students in the class originally.

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4 years ago

Read 2 more answers

What are he square roots of 36/100?

tiny-mole [99]

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3 years ago

A Survey of 85 company employees shows that the mean length of the Christmas vacation was 4.5 days, with a standard deviation of

GenaCL600 [577]

Answer:

The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).

The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).

Step-by-step explanation:

We have the standard deviations for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 85 - 1 = 84

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 84 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$ . So we have T = 1.989.

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.989\frac{1.2}{\sqrt{85}} = 0.26$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.26 = 4.24 days

The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.26 = 4.76 days

The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).

92% confidence interval:

Following the sample logic, the critical value is 1.772. So

$M = T\frac{s}{\sqrt{n}} = 1.772\frac{1.2}{\sqrt{85}} = 0.23$

The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.23 = 4.27 days

The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.23 = 4.73 days

The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).

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3 years ago

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3 years ago