(r/s)(3) = ?<br><br> please help.

Can y’all help me on question 19?!

m_a_m_a [10]

Answer: 178.31

Step-by-step explanation: subtract 574.54 by 396.23

8 0

3 years ago

Read 2 more answers

The structure of a two-factor study can be presented as a matrix with the levels of one factor determining the rows and the leve

slega [8]

Answer:

Following are the solution to this question:

Step-by-step explanation:

It provides three different hypotheses in such a two-factor ANOVA:

In point A:

H o: With all factor A levels, the ways are equivalent

Ha: A least another element A level does have a transfer to another

In point B:

Ha: The least one Factor A the level does have a transfer to another

H o: With all Factor B levels the results are about the same.

In point C:

Ha: At most one Variable B level does have a transfer than any other level.

H o: There are no interactions among the factors

Ha: The interactions of factors are important

When ANOVA is executed, they get three p-sets (one for all 3 hypotheses)

(a) If Variable A's p-value is much less alpha, we will reject the null and embrace Ha and infer that Factor A is important. Anything else, H o also isn't rejected and that there is no evidence which Factor A is important

(b) If p- is < alpha, otherwise we reject The null, accept Ha, and infer which Factor B is relevant. Factor B is significant. Conversely, we may not condemn H o but claim there isn't enough proof which Factor B is important

(c)

If they reject H o and agree to the point p- for the A x B interaction is a < alpha Ha, and conclude that the interaction from A to B is important. So, perhaps we can deny H o and claim, that neither proof of interactions is sufficient From A to B.

8 0

3 years ago

Express this number in scientific notation. 0.15880

Marysya12 [62]

Answer:

Step-by-step explanation:

3 0

3 years ago

Backgammon is a board game for two players in which the playing pieces are moved according to the roll of two dice. players win

Ivahew [28]

The rolls of the dice are independent, i.e. the outcome of the second die doesn't depend in any way on the outcome of the first die.

In cases like this, the probability of two events happening one after the other is the multiplication of the probabilities of the two events.

So, the probability of rolling two 6s is the multiplication of the probabilities of rolling a six with the first die, and another six with the second:

$P(\text{rolling two 6s}) = P(\text{rolling a 6}) \cdot P(\text{rolling a 6}) = \dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{36}$

Similarly,

$P(\text{rolling two 3s}) = P(\text{rolling a 3}) \cdot P(\text{rolling a 3}) = \dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{36}$

Actually, you can see that the probability of rolling any ordered couple is always 1/36, since the probability of rolling any number on both dice is 1/6:

$P(\text{rolling any ordered couple}) = P(\text{rolling the first number}) \cdot P(\text{rolling the second number}) = \dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{36}$