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3 years ago

8

The partially complete anova table given here is for a two-factor factorial experiment source df ss ms factor a 3 2.25 0.75 fact

or b 1 0.95 interaction 3 0.30 error 16 2.40 0.15 total 6.50

1 answer:

ycow [4]3 years ago

6 0

Its A I'm very sure

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8 more boys
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3 years ago

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U=10y/3x<br><br>Solve for x

SSSSS [86.1K]

Answer:

x=(3u)/(10y)

Step-by-step explanation:

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Determine which package of laundry soap is the better buy: 65 ounces for 6.99 of 120 ounces of 12.99

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3 years ago

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3 years ago

In 2013, the moose population in a park was measured to be 5,100. By 2018, the population was measured again to be 5,200. If the

natka813 [3]

Answer:

$P(t) = 5100e^{0.0039t}$

Step-by-step explanation:

The exponential model for the population in t years after 2013 is given by:

$P(t) = P(0)e^{rt}$

In which P(0) is the population in 2013 and r is the growth rate.

In 2013, the moose population in a park was measured to be 5,100

This means that $P(0) = 5100$

So

$P(t) = 5100e^{rt}$

By 2018, the population was measured again to be 5,200.

2018 is 2018-2013 = 5 years after 2013.

So this means that $P(5) = 5200$ .

We use this to find r.

$P(t) = 5100e^{rt}$

$5200 = 5100e^{5r}$

$e^{5r} = \frac{52}{51}$

$\ln{e^{5r}} = \ln{\frac{52}{51}}$

$5r = \ln{\frac{52}{51}}$

$r = \frac{\ln{\frac{52}{51}}}{5}$

$r = 0.0039$

So the equation for the moose population is:

$P(t) = 5100e^{0.0039t}$

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3 years ago