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3 years ago

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You have space in your garden for 60 small flowering bulbs. Crocus bulbs cost $0.25 each and daffodil bulbs cost $0.65 each. You

r budget allows you to spend $27.80 on bulbs. How many crocus bulbs and how many daffodil bulbs can you buy?

1 answer:

Doss [256]3 years ago

6 0

Answer:

32 daffodils bulbs and 28 crocus bulbs

Step-by-step explanation:

Assuming 'x' as Crocus and 'y' as daffodil. Now we'll form equations.

-> for 60 small flowering bulbs

x+y=60

x=60-y --->eq(1)

->Crocus bulbs cost $0.25 each and daffodil bulbs cost $0.65 each with the budget $27.80

0.25x + 0.65y = 27.80 -->eq(2)

Next is to substitute x=60-y in eq(2)

(2)=>0.25(60-y) + 0.65y = 27.80

15 - 0.25y + 0.65y = 27.80

0.4y = 12.8

y = 12.8/ 0.4

y= 32 daffodil bulbs

Now, plugging the value of y in eq(1)

(1)=>x=60-32

x= 28

x=28 crocus bulbs

Therefore, you can buy 32 daffodils bulbs and 28 crocus bulbs

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The two nominal variables are related.

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Ownership C1 C2 C3

No 40 32 48

Yes 30 48 52

The first thing to do in order to determine if they are related or not is to state our null and alternative hypothesis

Null hypothesis

$\mathbf{H_o: Two \nomial \ variables \ are \ related}$

Alternative hypothesis

$\mathbf{H_a: Two \nomial \ variables \ are \ not \ related}$

Using the Chi-square test statistics which can be expressed by using the formula

$X ^2 = \sum \dfrac{(O-E)^2}{E}$

Ownership C1 C2 C3 Total

No 40 32 48 120

Yes 30 48 52 130

Total 70 80 100 250

The expected values are calculated as:

$\mathsf{E_{a,b} = \dfrac{(row \ total \times column \ total )}{grand \ total }}$

$\mathsf{E_{1,1} = \dfrac{(70 \times120 )}{250 }}$

$\mathsf{E_{1,1} = 33.6}$

$\mathsf{E_{1,2} = \dfrac{(70 \times130 )}{250 }}$

$\mathsf{E_{1,2} = 36.4}$

$\mathsf{E_{2,1} = \dfrac{(80 \times120 )}{250 }}$

$\mathsf{E_{2,1} = 38.4}$

$\mathsf{E_{2,2} = \dfrac{(80 \times 130 )}{250 }}$

$\mathsf{E_{2,2} = 41.6}$

$\mathsf{E_{3,1} = \dfrac{(100 \times120 )}{250 }}$

$\mathsf{E_{3,1} = 48}$

$\mathsf{E_{3,2} = \dfrac{(70 \times130 )}{250 }}$

$\mathsf{E_{3,2} = 52}$

∴ Using the Chi-square test statistics, we have:

$X ^2 = \sum \dfrac{(O-E)^2}{E}$

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$X ^2 = \Bigg( \dfrac{40.96}{33.6}+ \dfrac{40.96}{36.4}+ \dfrac{40.96}{38.4}+ \dfrac{40.96}{41.6} + \dfrac{0}{48}+ \dfrac{0}{52} \Bigg)$

$X ^2 = \Bigg( 1.2190+ 1.1253+ 1.0667+ 0.9846+0+ 0 \Bigg)$

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The degree of freedom df = ((r - 1) × (c - 1))

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∴

Assuming the level of significance = 5%

The p-value of the Chi-square test statistics at df of 2 is:

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Therefore, we can conclude that since the p-value (0.111) is greater than the level of significance (0.05), we fail to reject the null hypothesis.

Hence, the two nominal variables are related.

Learn more about Chi-square test statistics here:

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