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3 years ago

7

Every week, Mr. Kirkson uses 3 1/6 gallon of water to water every 1/3 square foot of his garden. How many gallons of water does

Mr. Kirkson use per square foot to water his garden each week?

2 answers:

Nikolay [14]3 years ago

7 0

Answer:

Mr. Kirkson uses 9.60 gallons of water per square foot to water his garden each week.

Step-by-step explanation:

This problem can be solved by a simple rule of three.

We have that 1/3 = 0.33 square foot of his garden uses uses 3 1/6 = 19/6 = 3.167 gallons of water. How many are used per square foot? So:

1 foot - x gallons

0.33 foot - 3.167 gallons

$0.33x = 3.167$

$x = \frac{3.167}{0.33}$

$x = 9.60$

Mr. Kirkson uses 9.60 gallons of water per square foot to water his garden each week.

nata0808 [166]3 years ago

5 0

3 1/6 / 1/3 =

19/6 / 1/3 =

19/6 * 3/1 = 57/6 = 9 1/2 gallons per square foot

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Answer:

Step-by-step explanation:

The summary of the given data includes;

sample size for the first school $n_1$ = 42

sample size for the second school $n_2$ = 34

so 16 out of 42 i.e $x_1$ = 16 and 18 out of 34 i.e $x_2$ = 18 have ear infection.

the proportion of students with ear infection Is as follows:

$\hat p_1 = \dfrac{16}{42}$ = 0.38095

$\hat p_2 = \dfrac{18}{34}$ = 0.5294

Since this is a two tailed test , the null and the alternative hypothesis can be computed as :

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level of significance ∝ = 0.05,

Using the table of standard normal distribution, the value of z that corresponds to the two-tailed probability 0.05 is 1.96. Thus, we will reject the null hypothesis if the value of the test statistics is less than -1.96 or more than 1.96.

The test statistics for the difference in proportion can be achieved by using a pooled sample proportion.

$\bar p = \dfrac{x_1 +x_2}{n_1 +n_2}$

$\bar p = \dfrac{16 +18}{42 +34}$

$\bar p = \dfrac{34}{76}$

$\bar p = 0.447368$

$\bar p + \bar q = 1 \\ \\ \bar q = 1 -\bar p \\ \\\bar q = 1 - 0.447368 \\ \\\bar q = 0.552632$

The pooled standard error can be computed by using the formula:

$S.E = \sqrt{ \dfrac{ \bar p \bar q}{ n_1} + \dfrac{\bar p \bar p}{n_2} }$

$S.E = \sqrt{ \dfrac{ 0.447368 * 0.552632}{ 42} + \dfrac{ 0.447368 * 0.447368}{34} }$

$S.E = \sqrt{ \dfrac{ 0.2472298726}{ 42} + \dfrac{ 0.2001381274}{34} }$

$S.E = \sqrt{ 0.01177284105}$

$S.E = 0.1085$

The test statistics is ;

$z = \dfrac{\hat p_1 - \hat p_2}{S.E}$

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$z = \dfrac{-0.14845}{0.1085}$

z = - 1.368

Decision Rule: Since the test statistics is greater than the rejection region - 1.96 , we fail to reject the null hypothesis.

Conclusion: There is insufficient evidence to support the claim that a difference exists between the proportions of students who have ear infections at the two schools

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