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saveliy_v [14]
3 years ago
11

PLEASE HELP ME WITH THIS SOMEONE

Mathematics
2 answers:
xxTIMURxx [149]3 years ago
8 0

Answer:

try d

Step-by-step explanation:

because if you look at what point a is it you get 9 and if you look at where point b is at you get 9,3 but pint b says 2 so im thinking it was publish error on that one so the answer will be 9,3 have a good day and stay safe

Simora [160]3 years ago
7 0
Try d because i think it’s d
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Which of the following conditions must be met in order to make a statistical inference about a population based on a sample
klasskru [66]

Answer:

For this case if we want to conclude that  the sample does not come from a normally distributed population we need to satisfy the condition that the sample size would be large enough in order to use the central limit theoream and approximate the sample mean with the following distribution:

\bar X \sim (\mu, \frac{\sigma}{\sqrt{n}})

For this case the condition required in order to consider a sample size large is that n>30, then the best solution would be:

n>= 30

Step-by-step explanation:

For this case if we want to conclude that  the sample does not come from a normally distributed population we need to satisfy the condition that the sample size would be large enough in order to use the central limit theoream and approximate the sample mean with the following distribution:

\bar X \sim (\mu, \frac{\sigma}{\sqrt{n}})

For this case the condition required in order to consider a sample size large is that n>30, then the best solution would be:

n>= 30

7 0
3 years ago
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

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