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Mashcka [7]
3 years ago
7

Denise wants to cover a 21.75 square foot wall with weathered wood. If each box of weathered wood contains 1.5 square feet of wo

od, how many boxes of weathered wood does she need?
Mathematics
1 answer:
natka813 [3]3 years ago
6 0

Answer: 15 boxes

Step-by-step explanation:

From the question, we are informed that Denise wants to cover a 21.75 square foot wall with weathered wood and that each box of weathered wood contains 1.5 square feet of wood.

The number of boxes of weathered wood that she need will be calculated by dividing 21.75 square feet by 1.5 square feet. This will be:

= 21.75/1.5

= 14.5

= 15 boxes approximately

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Answer:

Step-by-step explanation:

Corresponding heights of presidents and height of their main opponents form matched pairs.

The data for the test are the differences between the heights.

μd = the​ president's height minus their main​ opponent's height.

President's height. main opp diff

191. 166. 25

180. 179. 1

180. 168. 12

182. 183. - 1

197. 194. 3

180. 186. - 6

Sample mean, xd

= (25 + 1 + 12 - 1 + 3 + 6)/6 = 5.67

xd = 5.67

Standard deviation = √(summation(x - mean)²/n

n = 6

Summation(x - mean)² = (25 - 5.67)^2 + (1 - 5.67)^2 + (12 - 5.67)^2+ (- 1 - 5.67)^2 + (3 - 5.67)^2 + (- 6 - 5.67)^2 = 623.3334

Standard deviation = √(623.3334/6 sd = 10.19

For the null hypothesis

H0: μd ≥ 0

For the alternative hypothesis

H1: μd < 0

The distribution is a students t. Therefore, degree of freedom, df = n - 1 = 6 - 1 = 5

The formula for determining the test statistic is

t = (xd - μd)/(sd/√n)

t = (5.67 - 0)/(10.19/√6)

t = 1.36

We would determine the probability value by using the t test calculator.

p = 0.12

Since alpha, 0.05 < than the p value, 0.12, then we would fail to reject the null hypothesis.

Therefore, at 5% significance level, we can conclude that for the population of heights for presidents and their main​ opponents, the differences have a mean greater than 0 cm.

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Hello!

\large\boxed{c = 2}

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