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vlada-n [284]
4 years ago
9

Determine whether the statement describes a descriptive or inferential statistic. A recent poll of 1971 home owners in West Virg

inia showed that the average price of a house in the U.S. is $319,000.
Mathematics
1 answer:
allsm [11]4 years ago
8 0

Answer:

Inferential statistics

Step-by-step explanation:

When we conduct a poll, we aim to make a generalization about the population or units under study. The recent poll of 1971 home owners in West Virginia showed that the average (mean) price of house <em>in the U.S</em><em>.</em> is $319,000.

Recall that this poll was conducted only in West Virginia but the result or finding of the poll is being generalized over the U.S. as a whole. Hence, we are using the sample (fraction) to determine the population (whole) of home owners in the U.S. This is what we call <em>statistical inference or inferential statistics.</em>

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Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th
trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

    t = 20: 999 fishes

    t = 30: 1168 fishes

(c)

P(\infty) = 1200

Step-by-step explanation:

Given

P(t) =\frac{d}{1+ke^-{ct}}

d = 1200\\k = 11\\c=0.2

Solving (a): Fishes at t = 0

This gives:

P(0) =\frac{1200}{1+11*e^-{0.2*0}}

P(0) =\frac{1200}{1+11*e^-{0}}

P(0) =\frac{1200}{1+11*1}

P(0) =\frac{1200}{1+11}

P(0) =\frac{1200}{12}

P(0) = 100

Solving (a): Fishes at t = 10, 20, 30

t = 10

P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483

t = 20

P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999

t = 30

P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168

Solving (c): \lim_{t \to \infty} P(t)

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of e^{-ct} decreases

This implies that:

{t \to \infty} = {e^{-ct} \to 0}

So:

The value of P(t) for large values is:

P(\infty) = \frac{1200}{1 + 11 * 0}

P(\infty) = \frac{1200}{1 + 0}

P(\infty) = \frac{1200}{1}

P(\infty) = 1200

5 0
3 years ago
Sammy Sales had the following receipts for his tree trimming business for the first six months of the year. What is the average
fredd [130]

average =  \frac{sum \:  \: of \:  \:all \:  \:  obeservations}{no. \:  \: of \: observaions}

=  \frac{1205 + 1450 + 1530 + 1655 + 1610 + 1710}{6}

=  \frac{9160}{6}

=  1526.66

___________________________________________

1526.66 is the average of his sales for this period.

6 0
3 years ago
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