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spin [16.1K]
3 years ago
15

The perimeter of a rectangle is 68 ft. The ratio of the length to the width is 9 : 8. What are the dimensions of the rectangle?

Mathematics
2 answers:
IceJOKER [234]3 years ago
8 0

The answer to your question is... 9x + 9x + 8x + 8x = 68

Hope this helps ;) I had the test too and it was right

likoan [24]3 years ago
5 0
18 foot by 16 foot hope this helps
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The answer is 1/3...I got this question on I-ready and I hate when I don't pass a lesson...hope this will help you guys out.
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The course grade in a statistics class is the average of the scores on five examinations. Suppose that a student's scores on the
nadya68 [22]

Answer:

84 is the highest possible course average

Step-by-step explanation:

Total number of examinations = 5

Average = sum of scores in each examination/total number of examinations

Let the score for the last examination be x.

Average = (66+78+94+83+x)/5 = y

5y = 321+x

x = 5y -321

If y = 6, x = 5×6 -321 =-291.the student cannot score -291

If y = 80, x = 5×80 -321 =79.he can still score higher

If If y = 84, x = 5×84 -321 =99.This would be the highest possible course average after the last examination.

If y= 100

The average cannot be 100 as student cannot score 179(maximum score is 100)

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2 years ago
Use the table to interpret the axes labels of the
yKpoI14uk [10]

Answer:

Texting speed

Step-by-step explanation:

7 0
3 years ago
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Question in attachment
kiruha [24]

Answer:

1. a= 7, A = 49 degrees

Side A = 7 , Side B = 24, Side C = 25

A= 49 degrees, C = 90 degrees , B=41

2. A= 4, B = 6.9, C=8

A= 22 degrees, B = 68, C= 90

3. A= 7, B = 14.4, C=16

A= 45 , B=45  , C=90

6 0
3 years ago
A sample of 1200 computer chips revealed that 45% of the chips fail in the first 1000 hours of their use. The company's promotio
yaroslaw [1]

Answer:

z=\frac{0.45 -0.48}{\sqrt{\frac{0.48(1-0.48)}{1200}}}=-2.08

p_v = P(Z

So the p value obtained was a low value and using the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of chips that fail in the first 1000 hours of their use is not significantly less than 0.48.   

Step-by-step explanation:

Data given and notation

n=1200 represent the random sample taken

\hat p=0.45 estimated proportion of chips that fail in the first 1000 hours of their use

\mu_0 =0.48 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

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 the true proportion si less then 0.48:  

Null hypothesis:p\geq 0.48  

Alternative hypothesis:p < 0.48  

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  is significantly different from a hypothesized value .

Calculate the statistic  

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

z=\frac{0.45 -0.48}{\sqrt{\frac{0.48(1-0.48)}{1200}}}=-2.08

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 significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

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

p_v = P(Z

So the p value obtained was a low value and using the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of chips that fail in the first 1000 hours of their use is not significantly less than 0.48.  

6 0
3 years ago
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