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3 years ago

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You are a contractor who is installing a new wooden fence for a customer. The dimensions of the fence are 20 feet long and 6 fee

t high. If each piece of wood is 72 inches tall and 4 inches wide and the wood costs $1.10 per square foot, how much will the wood for the new fence cost?

1 answer:

ahrayia [7]3 years ago

7 0

Answer:

$132

Step-by-step explanation:

First we need to find the total number of woods required. This would be total area of fence divided by area of each wood. Remember to convert inch into feet.

$\frac{area of fence}{area of wood} \\=\frac{20 * 6}{6 * 0.3333} \\= 60 woods plank$

Now we know the total number of planks, we need to find what is the cost of each plank. We know that one wood plank area is 2 sq. ft ( 6*0.33333). Therefore, the cost of each plank is = cost per area(sq.ft) multiply by total area (sq. ft).

cost per plank = 2 * $1.10 =$2.20.

We have total planks = 60 units and we have cost per plank = $2.20.

So, we can find total cost as 60*2.20= $132

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<em>C₁ </em>: <em>x</em> = <em>t</em> and <em>y</em> = <em>t</em> ² with 0 ≤ <em>t</em> ≤ 1

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Step-by-step explanation:

1) Data given and notation

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Assuming these values for the remaining data:

$\sigma_{1}=1.2$ represent the population standard deviation for the sample men

$\sigma_{2}=1.5$ represent the population standard deviation for the sample women

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$n_{2}=30$ sample size for the group women

z would represent the statistic (variable of interest)

$p_v$ represent the p value

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for men it's higher than the mean for women, the system of hypothesis would be:

H0: $\mu_{1} \leq \mu_{2}$

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If we analyze the size for the samples both are higher than 30, and we know the population deviation's, so for this case is better apply a z test to compare means, and the statistic is given by:

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We have all in order to replace in formula (1) like this:

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In order to find the critical value we need to take in count that we are conducting a right tailed test, so we are looking on the normal standard distribution a value that accumulats 0.01 of the area on the right and 0.99 of the area on the left. We can us excel or a table to find it, for example the code in Excel is:

"=NORM.INV(1-0.01,0,1)", and we got $z_{critical}=2.33$

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Since our calculated value is lower than our critical value, $z_{calc}=2.02$ , we have enough evidence to FAIL to reject the null hypothesis at 1% of significance. So there is not nough evidence to conclude that mean for men it's significantly higher than the mean for female.

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