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Westkost [7]
3 years ago
13

The assembly room has 6 rows with 30 chair in each row. If third grader fill 3 rows, how many third grader are in the room?

Mathematics
2 answers:
mr Goodwill [35]3 years ago
7 0
30 x 3 = 90 their are 90 3rd graders in the room
Anuta_ua [19.1K]3 years ago
7 0
30/6=5 chairs each row
5x3=15 chairs. Meaning that there are 15 third graders in the room.
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7 plus 9.  10 plus 6 gives you 16, as does 7 plus 9.
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Determine whether the given value is a statistic or a parameter. A homeowner measured the voltage supplied to his home on a rand
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5 but its probaly not right bc im just looking for pointsStep-by-step explanation:

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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
In a random sample of 300 attendees of a minor league baseball game, 182 said that they bought food from the concession stand. C
horrorfan [7]

Answer:

<em>90% confidence interval for the proportion of fans who bought food from the concession stand</em>

(0.5603,0.6529)

Step-by-step explanation:

<em>Step(i)</em>:-

<em>Given sample size 'n' =300</em>

Given data  random sample of 300 attendees of a minor league baseball game, 182 said that they bought food from the concession stand.

<em>Given sample proportion </em>

<em>                        </em>p^{-}  = \frac{x}{n} = \frac{182}{300} =0.606

level of significance = 90% or 0.10

Z₀.₁₀ = 1.645

<em>90% confidence interval for the proportion is determined by</em>

(p^{-} - Z_{0.10}\sqrt{\frac{p(1-p)}{n} }  , p^{-} +Z_{0.10}\sqrt{\frac{p(1-p)}{n} })

(0.6066 - 1.645\sqrt{\frac{0.6066(1-0.6066)}{300} }  ,0.6066+1.645\sqrt{\frac{0.6066(1-0.6066)}{300} })

(0.6066 -  0.0463  ,0.6066 +  0.0463)

(0.5603,0.6529)

<u>final answer</u>:-

<em>90% confidence interval for the proportion of fans who bought food from the concession stand</em>

(0.5603,0.6529)

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For the polynomial below, what is the coefficient of the term with the power of 3?
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The answer is
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