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UNO [17]
2 years ago
12

Which are equivalent expressions?

Mathematics
1 answer:
bonufazy [111]2 years ago
6 0

Answer:

B) 4x + 8y = 2(2x + 4y)

C) 4x + 8y = 4(x + 2y)

D) 4x + 8y + 12z = 4(x + 2y + 3z)

E) 5x + 10y = 5(x + 2y)

Step-by-step explanation:

We would assume a value for the variables a, b, c, x, y and z.

Let x, a = 1

Let y, b = 2

Let z, c = 3

A) a + b + c = 3abc

When a = 1, b = 2 and c = 3

Substituting into the above equation, we have;

1 + 2 + 3 = 3(1*2*3)

6 ≠ 3(6)

6 ≠ 18 (not equivalent)

B) 4x + 8y = 2(2x + 4y)

When x = 1 and y = 2

Substituting into the above equation, we have;

4(1) + 8(2) = 2(2*1 + 4*2)

4 + 16 = 2(2 + 8)

20 = 2(10)

20 = 20 (equivalent expression)

C) 4x + 8y = 4(x + 2y)

When x = 1 and y = 2

Substituting into the above equation, we have;

4(1) + 8(2) = 4(1 + 2*2)

4 + 16 = 4(1 + 4)

20 = 4(5)

20 = 20 (equivalent expression).

D) 4x + 8y + 12z = 4(x + 2y + 3z)

When x = 1, y = 2 and z = 3

Substituting into the above equation, we have;

4(1) + 8(2) + 12(3) = 4(1 + 2*2 + 3*3)

4 + 16 + 36 = 4(1 + 4 + 9)

56 = 4(14)

56 = 56 (equivalent expression).

E) 5x + 10y = 5(x + 2y)

When x = 1 and y = 2

Substituting into the above equation, we have;

5(1) + 10(2) = 5(1 + 2*2)

5 + 20 = 5(1 + 4)

25 = 5(5)

25 = 25 (equivalent expression).

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Answer:

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Step-by-step explanation:

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Assume that a cross is made between a heterozygous tall pea plant and a homozygous short pea plant. Fifty offspring are produced
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Answer:

1) \chi^2 = \frac{(30-25)^2}{25} +\frac{(20-25)^2}{25} =1 +1 =2

2) df = c-1 = 2-1

Where c represent the number of categories c=2

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Tall =30 , Short =20

We need to conduct a chi square test in order to check the following hypothesis:

H0: The deviations from a 1:1 ratio (25 tall and 25 short) are due to chance.

H1: The deviations from a 1:1 ratio (25 tall and 25 short) are NOT due to chance.

Part 1

So then we know that the expected values would be 25 for each case

The statistic to check the hypothesis is given by:

\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}

And if we replace we got:

\chi^2 = \frac{(30-25)^2}{25} +\frac{(20-25)^2}{25} =1 +1 =2

Part 2

For this case the degreed of freedom are given by:

df = c-1 = 2-1

Where c represent the number of categories c=2

And we can calculate the p value given by:

p_v = P(\chi^2_{1} >2)=0.157

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(2,1,TRUE)"

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