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bearhunter [10]
3 years ago
15

The annual tuition at a specific college was $20,500 in 2000, and $45,4120

Mathematics
1 answer:
nika2105 [10]3 years ago
6 0

Answer: the tuition in 2020 is $502300

Step-by-step explanation:

The annual tuition at a specific college was $20,500 in 2000, and $45,4120 in 2018. Let us assume that the rate of increase is linear. Therefore, the fees in increasing in an arithmetic progression.

The formula for determining the nth term of an arithmetic sequence is expressed as

Tn = a + (n - 1)d

Where

a represents the first term of the sequence.

d represents the common difference.

n represents the number of terms in the sequence.

From the information given,

a = $20500

The fee in 2018 is the 19th term of the sequence. Therefore,

T19 = $45,4120

n = 19

Therefore,

454120 = 20500 + (19 - 1) d

454120 - 20500 = 19d

18d = 433620

d = 24090

Therefore, an

equation that can be used to find the tuition y for x years after 2000 is

y = 20500 + 24090(x - 1)

Therefore, at 2020,

n = 21

y = 20500 + 24090(21 - 1)

y = 20500 + 481800

y = $502300

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Step-by-step explanation:

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z_{\alpha/2} represent a quantile of the normal standard distribution that accumulates {\alpha/2} on each tail of the distribution.

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Squaring both sides:

(\frac{ME}{z_{\alpha/2}})^2=\frac{\hat p(1-\hat p)}{n}

And solving for n we got:

n=\frac{\hat p(1-\hat p)}{(\frac{ME}{z_{\alpha/2}})^2}

Now we can replpace the values

n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.644854})^2}=422.739

And rounded up to the nearest integer we got:

n=423

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