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Marta_Voda [28]
3 years ago
7

3 times 5 with a two on the top

Mathematics
1 answer:
jenyasd209 [6]3 years ago
3 0
Ok so it could be
1. \frac{2}{3 times 5}
2. 3 timies \frac{2}{5}


ok if 1. then
\frac{2}{3 times 5}=\frac{2}{15}

if 2.  3 timies \frac{2}{5}= \frac{6}{5}



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Which statement is false?
Zarrin [17]
<span>Let's analyze our choices:
1. Media messages may translate differently across different media. Let's see here, if a person is reading a newspaper, would they react differently than if they noticed a tweet by their friend online about presidential campaign ads? Yes, they probably would, so this statement is true.
2. When analyzing media, it is important to ask, "Does it work ". If a person told you that the sky is actually pink but Mary Poppins is actually a real person and just makes you think that the sky looks blue, and that they learned this because the news told them so, would you automatically believe them? Not really. :P Therefore option 2 is out of the way.
3. It is important to understand that media messages do not have a goal. HAH! People and messages always have a goal. I think we may have found our false statement, but just to be sure, let's take a look at our last statement.
4. People will perceive media messages differently. </span>Have you ever watched a debate and thought that one side did a better job than the other, and then your friend starts an argument with you as they think that's total baloney? This statement is true too.

This leaves the only false statement as 3. <span>It is important to understand that media messages do not have a goal.</span>

Did that answer your question?
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

5 0
3 years ago
8:24 in simplest form
Mamont248 [21]

Answer:

1/3

Step-by-step explanation:

4 0
3 years ago
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I think that the answer to the question is 0.75
7 0
3 years ago
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THE ANSWER FOR THIS IS G. 16
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