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FrozenT [24]
3 years ago
13

an ecosystem which is 10 km squared contains a population of 1000 squirrels what is the population density of the squirrels in t

his ecosystem
Mathematics
1 answer:
Anastasy [175]3 years ago
4 0

Answer:

100 squirrels/km^2

Step-by-step explanation:

we know that

To find the population density of the squirrels in the ecosystem, divided the quantity of population of squirrels by the area of the ecosystem

so

1,000/10=100 squirrels/km^2

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Karl's car went 550 miles on 33 gallons of gas. Which equation does NOT give the correct unit rate, x, in miles per gallon for K
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x = 550 ÷ 33




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Find the least common multiple of 6n and 8c
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(Based on Q1 ~ Q3) According to the Bureau of the Census, 18.1% of the U.S. population lives in the Northeast, 21.9% inn the Mid
vekshin1

Answer:

We can therefore conclude that the geographical distribution of hotline callers could be the same as the U.S population distribution.

Step-by-step explanation:

The null Hypothesis: Geographical distribution of hotline callers could be the same as the U.S. population distribution

Alternative hypothesis: Geographical distribution of hotline callers could not be the same as the U.S. population distribution

The populations considered are the Midwest, South, Northeast, and west.

The number of categories, k = 4

Number of recent calls = 200

Let the number of estimated parameters that must be estimated, m = 0

The degree of freedom is given by the formula:

df = k - 1-m

df = 4 -1 - 0 = 3

Let the significance level be, α = 5% = 0.05

For  α = 0.05, and df = 3,

from the chi square distribution table, the critical value = 7.815

<u>Observed and expected frequencies of calls for each of the region:</u>

<u>Northeast</u>

Observed frequency = 39

It contains 18.1% of the US Population

The probability = 0.181

Expected frequency of call = 0.181 * 200 = 36.2

<u>Midwest</u>

Observed frequency = 55

It contains 21.9% of the US Population

The probability = 0.219

Expected frequency of call = 0.219 * 200 =43.8

<u>South</u>

Observed frequency = 60

It contains 36.7% of the US Population

The probability = 0.367

Expected frequency of call = 0.367 * 200 = 73.4

<u>West</u>

Observed frequency = 46

It contains 23.3% of the US Population

The probability = 0.233

Expected frequency of call = 0.233 * 200 = 46

x^{2} = \sum \frac{(O_{i} - E_{i})  ^{2} }{E_{i} } ,   i = 1, 2,.........k

Where O_{i} = observed frequency

E_{i} = Expected frequency

Calculate the test statistic value, x²

x^{2} = \frac{(39 - 36.2)^{2} }{36.2} + \frac{(55 - 43.8)^{2} }{43.8} + \frac{(60 - 73.4)^{2} }{73.4} + \frac{(46 - 46.6)^{2} }{46.6}

x^{2} = 5.535

Since the test statistic value, x²= 5.535 is less than the critical value = 7.815, the null hypothesis will not be rejected, i.e. it will be accepted. We can therefore conclude that the geographical distribution of hotline callers could be the same as the U.S population distribution.  

7 0
3 years ago
Help me with number 1 I don't get it
IgorLugansk [536]
The answer will be 250 because you divide 2 by 500
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3 years ago
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