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hodyreva [135]
3 years ago
15

Ms. Rao buys a computer, a printer, and a scanner for $2,543. The computer costs $1,502 more than the printer. The printer costs

$123 more than the scanner. How much does Ms. Rao pay for the computer?
Mathematics
1 answer:
Neporo4naja [7]3 years ago
5 0
Ms rao paid $1379for the computer.
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Yea thank for the answer
kolbaska11 [484]

Answer:

Step-by-step explanation:

Hello!

10*17 = > 170 cm²

(3,14 * 8)/4 =>6,28cm²

170 + 6,28 => 176,28cm²

4 0
2 years ago
Read 2 more answers
In the last three years frederico's basketball team won 40 more games than they lost if they won 150 games what was their ratio
Westkost [7]
150 games won: 110 games lost

Ratios
150:110
150/110
150 to 110
3 0
3 years ago
The enrollment at a local college increased 2% over last year's enrollment of 3300. Find the current enrollment.
ella [17]
It should be 3366. To find the percent increased, multiply 3300 by .02 and add that to 3300, giving you 3366.
8 0
3 years ago
Lucy si thinking of a Numbers. The number is greater than two hundred twenty-five. Her number is less than 2 hundreds,2 tens, an
VikaD [51]
226 i believe, if not then idk
8 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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