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3 years ago

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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10+ 6 has the same sum as 7+

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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:

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

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

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

$P(\infty) = 1200$

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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:

90% confidence interval for the proportion of fans who bought food from the concession stand

(0.5603,0.6529)

Step-by-step explanation:

Step(i):-

Given sample size 'n' =300

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

Given sample proportion

$p^{-} = \frac{x}{n} = \frac{182}{300} =0.606$

level of significance = 90% or 0.10

Z₀.₁₀ = 1.645

90% confidence interval for the proportion is determined by

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

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

(0.6066 - 0.0463 ,0.6066 + 0.0463)

(0.5603,0.6529)

final answer:-

90% confidence interval for the proportion of fans who bought food from the concession stand

(0.5603,0.6529)

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3 years ago

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

Triss [41]

The answer is
D.1

because even though there is not a number in front of X, we are to assume it is 1.

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