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babymother [125]
2 years ago
13

When we reject the null hypothesis, we would expect the variance ratio, over the long run, to be:_________

Mathematics
1 answer:
FinnZ [79.3K]2 years ago
5 0

Answer:

Cuando rechazamos la hipótesis nula, esperaríamos que la razón de varianza, a largo plazo, sea: <u>mayor al valor crítico.</u>

Step-by-step explanation:

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Find the probability of rolling an odd sum or a sum less than 7 when a pair of dice is rolled
Nutka1998 [239]

Answer:

For the pairs that satisfy that the sum is less than 7 we have:

(1,1) (2,1) (1,2) (3,1) (2,2) (1,3) (4,1) (3,2) (2,3) (1,4) (5,1) (4.2) (3,3) (2,4) (1,5)

So we have a total of 15 pairs where the sum is less than 7

And for an odd sum we have the following pairs

(2,1) (1,2) (4,1) (3,2) (2,3) (1,4) (6,2) (5,2) (4,3) (3,4) (2,5) (1,6) (6,3) (5,4) (4,5) (3,6) (6,5) (5,6)

A total of 18 pairs and we have 6 pairs who are in both cases for the sum less than 7 and the sum an odd number that represent the intersection and using the total probability rule we got:

P = \frac{15+18-6}{36}= \frac{27}{33}= 0.818

Step-by-step explanation:

For this case when a pair of dice is rolled we have the following sample pace for the outcomes:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

We see 36 possible outcomes and we want to find how many of these pairs we got rolling an odd sum or a sum less than 7

For the pairs that satisfy that the sum is less than 7 we have:

(1,1) (2,1) (1,2) (3,1) (2,2) (1,3) (4,1) (3,2) (2,3) (1,4) (5,1) (4.2) (3,3) (2,4) (1,5)

So we have a total of 15 pairs where the sum is less than 7

And for an odd sum we have the following pairs

(2,1) (1,2) (4,1) (3,2) (2,3) (1,4) (6,2) (5,2) (4,3) (3,4) (2,5) (1,6) (6,3) (5,4) (4,5) (3,6) (6,5) (5,6)

A total of 18 pairs and we have 6 pairs who are in both cases for the sum less than 7 and the sum an odd number that represent the intersection and using the total probability rule we got:

P = \frac{15+18-6}{36}= \frac{27}{33}= 0.818

7 0
3 years ago
20 POINTS!!!! will give more if verified
DaniilM [7]

Answer:

The bottom one. The side with the 12 x 5 side.

Step-by-step explanation:

12 x 5 =60. That is bigger than the other sides

4 0
2 years ago
5a + 4b^2<br> (if a = -11 and b = 3)
Degger [83]
5(-11)+4(3)^2 = -19
The answer is -19
8 0
3 years ago
Please help me with this ​
MissTica

Answer:

For the question on the right, I assume b is the exponential factor, or growth factor.

Step-by-step explanation:

Linear:

Y=aX + b\\(X, Y) = (0, 15) => b = 15\\(X,Y) = (1, 12) => a = -3\\Y = -3X + 15

You can find the equation by using two points and the equation for a line.

Exponential:

Y = a^{bX+c}\\(X, Y) = (0, 81) => a^{c} = 81 => a^{c} = 3^{4} => a=3, c=4\\(X, Y) = (1, 27) => a^{b+c} = 27 => a^{b}a^{c} = 27 => a^{b} = \frac{27}{81} = \frac{1}{3} => b=-1\\Y = 3^{-X+4}

6 0
2 years ago
What is the surface area of this figure?
disa [49]

Answer:

72 sq. mi

Step-by-step explanation:

Breaking this down, we have 2 right triangles with sides of 3, 4, and 5 miles, and 3 rectangles with dimensions 3 x 5, 4 x 5, and 5 x 5 miles. Remember that the area of a triangle is 1/2 x b x h , where b and h are the triangle's base and height. The base and height of the triangles at the bases of the figure are 3 and 4, so each triangle has an area of 1/2 x 3 x 4 = 1/2 x 12 = 6 sq. mi, or 6 + 6 = 12 sq. mi together.

Onto the rectangles, we can find their area by multiplying their length by their width. Since the width of these rectangles is the same for all three - 5 mi - we can make our lives a little easier and just "glue" the lengths together, giving us a longer rectangle with a length of 3 + 4 + 5 = 12 mi. Multiplying the two, we find the area of the rectangles to be 5 x 12 = 60 sq. mi.

Adding this area to the triangle's area gives us a total area of 12 + 60 = 72 sq. mi.

7 0
3 years ago
Read 2 more answers
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