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devlian [24]
3 years ago
7

An average swimming pool contains 19200 gallons of water. The pool needs drained at the end of the summer. Water drains from the

pool at the rate of 12 gallons per minute.
Which equation can be used to find the volume, V, of the water in the pool m minutes after draining begins?

Mathematics
1 answer:
erastovalidia [21]3 years ago
5 0

An average swimming pool contains 19200 gallons of water. The pool needs drained at the end of the summer. Water drains from the pool at the rate of 12 gallons per minute.

Which equation can be used to find the volume, V, of the water in the pool m minutes after draining begins?An average swimming pool contains 19200 gallons of water. The pool needs drained at the end of the summer. Water drains from the pool at the rate of 12 gallons per minute.

Which equation can be used to find the volume, V, of the water in the pool m minutes after draining begins?An average swimming pool contains 19200 gallons of water. The pool needs drained at the end of the summer. Water drains from the pool at the rate of 12 gallons per minute.

Which equation can be used to find the volume, V, of the water in the pool m minutes after draining begins?An average swimming pool contains 19200 gallons of water. The pool needs drained at the end of the summer. Water drains from the pool at the rate of 12 gallons per minute.

Which equation can be used to find the volume, V, of the water in the pool m minutes after draining begins?An average swimming pool contains 19200 gallons of water. The pool needs drained at the end of the summer. Water drains from the pool at the rate of 12 gallons per minute.

Which equation can be used to find the volume, V, of the water in the pool m minutes after draining begins?

61y)6y \times 2 \div  \sqrt[ \geqslant  | | \frac{ >  >  <  =  {348)48)98)8(978). >  \\  | >  | \sqrt[m \: \% log( \beta  \tan( \\  log_{ log_{ \binom{ \\  \\ {}cbc log(\pi \beta  \sec( \cos( \tan( \sin( \infty e log( ln( log( ln( > .15 - 756 { >  | >  | <  \sqrt[ < 5x6y6)9x9yyyy \div  \frac{ \\ 2697930046y - xy >  \geqslant \:  \binom{ log( \binom{ \sec( \gamma  \\ }m{cx > 03y9)y \times ) }{?} ) }{?}  \times \frac{?}{?} }{?}  \times \frac{?}{?} ]{?} | | }^{?} ) ) ) ) ) ) ) ) ) }{?} }(?) }(?) ) ) ]{?} | | }^{?} }{?} | | ]{?}

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Step-by-step explanation:

Hello!

You have the information of 3 groups of people.

Group 1

n₁= 20

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Group 2

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Group 3

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1. To manually calculate the mean square between the groups you have to calculate the sum of square between conditions and divide it by the degrees of freedom.

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Sum Square B is:

∑ni(Ÿi - Ÿ..)²

Ÿi= sample mean of sample i ∀ i= 1,2,3

Ÿ..= general mean is the mean that results of all the groups together.

General mean:

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Mean square B= Sum Square B/Df B= 10.64/2= 5.32

2. The mean square error (MSE) is the estimation of the variance error (σ_{e}^2 → S_{e} ^{2}), you have to use the following formula:

Se²=<u> (n₁-1)S₁² + -(n₂-1)S₂² + (n₃-1)S₃²</u>

                        n₁+n₂+n₃-k

Se²=<u> 19*14.3 + 19*17.2 + 19*16.7 </u>= <u>  915.8   </u>  = 16.067

                 20+20+20-3                  57

DfE= N-k = 60-3= 57

3. To calculate the value of the statistic you have to divide the MSB by MSE

F= \frac{Mean square B}{Mean square E} = \frac{5.32}{16.067} = 0.33

4. P(F_{2; 57} ≤ F) = P(F_{2; 57} ≤ 0.33) = 0.28

I hope you have a SUPER day!

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strojnjashka [21]
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PROOF 

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2         | 14(0.9)² = 11.34
3         | 14(0.9)³ = 10.206
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