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blsea [12.9K]
2 years ago
10

The length of a rectangular piece of cardboard is 2 cm greater than its width. If the length and the width were each decreased b

y 1 cm, the area of the cardboard would be decreased by 27 cm². What are the dimensions of the original piece of cardboard?
Write and solve an equation to represent the problem.
Mathematics
1 answer:
Vladimir [108]2 years ago
8 0

9514 1404 393

Answer:

  13 cm wide;  15 cm long

Step-by-step explanation:

Let w represent the original width. Then the original length is (w+2) and the original area is w(w+2). After the decrease, the width is (w-1) and the length is (w+1). The decreased area is (w-1)(w+1). The difference between these areas is 27, so we have ...

  w(w+2) -(w-1)(w+1) = 27

 w^2 +2w -(w^2 -1) = 27

 2w +1 = 27

  w = (27-1)/2 = 13

The original piece of cardboard was 13 cm wide and 15 cm long.

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Divide 6⁄13 by 6⁄12 . <br><br> A. 1⁄12<br> B. 9⁄16<br> C. 12⁄13<br> D. 13⁄12
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A buyer went to the market to buy strawberries. He purchased 120 randomly selected strawberries from a vendor who claimed that n
mafiozo [28]

Answer:

z=\frac{0.333 -0.25}{\sqrt{\frac{0.25(1-0.25)}{120}}}=2.10  

p_v =P(z>2.10)=0.018  

At 5% of significance we can conclude that the true proportion of strawberries damage is higher than 0.25

Step-by-step explanation:

Data given and notation

n=120 represent the random sample taken

X=40 represent the number of strawberries damaged

\hat p=\frac{40}{120}=0.333 estimated proportion of strawberries damaged

p_o=0.25 is the value that we want to test

\alpha represent the significance level

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that no more than 25% of his total harvest of strawberries was damaged.:  

Null hypothesis:p\leq 0.25  

Alternative hypothesis:p > 0.25  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.333 -0.25}{\sqrt{\frac{0.25(1-0.25)}{120}}}=2.10  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

p_v =P(z>2.10)=0.018  

At 5% of significance we can conclude that the true proportion of strawberries damage is higher than 0.25

6 0
2 years ago
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