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Dafna11 [192]
3 years ago
5

A low P-Value in ANOVA table mean that the population mean of each treatment effects are different. the population mean of at le

ast one treatment effect are different. the population mean of at least one treatment effect are the same. the population mean of each treatment effects are the same.
Mathematics
1 answer:
Verizon [17]3 years ago
4 0

Answer:

The population mean of at least one treatment effect are different.

Step-by-step explanation:

An analysis of variance (ANOVA) is conducted in order to determine if there are significant differences between the values of the population mean with respect to the response variable for the domains that under the effects of different treatments. A low p-value leads to reject the null hypothesis of the following hypothesis system:

H_0: \mu_1=\mu_2=...=\mu_k\\\\\\H_a: \mu_i \neq\mu_j, for $ for some i, j between 1 and k.$

Rejecting H0 means that this hypothesis is false and, in turn, allows us to conclude that the population mean of one of the domains is different from the others.

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Answer:

breadth = 8, length = 32

Step-by-step explanation:

perimeter = 2 x (length x breadth)

                = 2 x ((24 + b) + b))

                = 2 x (24 + 2b)

                = 48 + 4 b

48 + 4b meters costs rupees 2400

each meter costs rupees 30

so,

30 x (48 + 4b) = 2400

1440 + 120b    = 2400

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3 years ago
)The histogram shows the number of gold medals won by different numbers of students at an art competition. Which of the followin
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Answer:

Eight students won 6, 7, or 8 gold medals.

Step-by-step explanation:

We see that the third bin is labelled 6 - 8. Since this is on the horizontal, or x, axis, this means that the students in this "bin" have won 6 - 8 medals, which is the same as saying winning 6, 7, or 8 medals.

Because this "bin" goes to 8, that means that there are 8 students who earned this many medals.

Thus, the answer is D.

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Uninhibited growth can be modeled by exponential functions other than​ A(t) ​=Upper A 0 e Superscript kt. For ​ example, if an i
laila [671]

The question is incomplete. Here is the complete question.

Uninhibited growth can be modeled by exponential functions other than A(t)=A_{0}e^{kt}. for example, if an initial population P₀ requires n units of time to triple, then the function P(t)=P_{0}(3)^{\frac{t}{n} } models the size of the population at time t. An insect population grows exponentially. Complete the parts a through d below.

a) If the population triples in 30 days, and 50 insects are present initially, write an exponential function of the form P(t)=P_{0}(3)^{\frac{t}{n} } that models the population.

b) What will the population be in 47 days?

c) When wil the population reach 750?

d) Express the model from part (a) in the form A(t)=A_{0}e^{kt}.

Answer: a) P(t)=50(3)^{\frac{t}{30} }

              b) P(t) = 280 insects

              c) t = 74 days

             d) A(t)=50e^{0.037t}

Step-by-step explanation:

a) n is time necessary to triple the population of insects, i.e., n = 30 and P₀ = 50. So, Exponential equation for growth is

P(t)=50(3)^{\frac{t}{30} }

b) In t = 47 days:

P(t)=50(3)^{\frac{t}{30} }

P(47)=50(3)^{\frac{47}{30} }

P(47)=50(3)^{1.567}

P(47) = 280

In 47 days, population of insects will be 280

c) P(t) = 750

750=50(3)^{\frac{t}{30} }

\frac{750}{50}=(3)^{\frac{t}{30} }

(3)^{\frac{t}{n} }=15

Using the property <u>Power</u> <u>Rule</u> of logarithm:

log(3)^{\frac{t}{30} }=log15

\frac{t}{30}log(3)=log15

t=\frac{log15}{log3} .30

t = 74

To reach a population of 750 insects, it will take 74 days

d) To express the population growth into the described form, determine the constant k, using the following:

A(t) = 3A₀ and t = 30

A(t)=A_{0}e^{kt}

3A_{0}=A_{0}e^{30k}

3=e^{30k}

Use Power Rule again:

ln3=ln(e^{30k})

ln3=30k

k=\frac{ln3}{30}

k = 0.037

Equation for exponential growth will be:

A(t)=50e^{0.037t}

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