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makkiz [27]
3 years ago
14

Three tanks,A,B,C are used to store oil. Tank A contains 'n' litres of oil. Tank B contains (n+150) litres of oil. Tank C is emp

ty. Oil is pumped from tank A and from tank B into tank C so that all three tanks contain the same amount of oil. 500 litres of oil are pumped from tank A into tank C. work out the value of 'n'.
Mathematics
1 answer:
Yuliya22 [10]3 years ago
6 0
500 divide by 300 is 150
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1/12
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5 divided by 585 5 ÷ 585 answer
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The answer is :
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zheka24 [161]

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What is the area of a rectangle of 5/9 m in length and 7/8 m in width
Andru [333]

Answer:

35/72 m^2

Step-by-step explanation:

Rectangle Formula: l*w=Area

Switch the variables with the numbers given:

\frac{5}{9} *\frac{7}{8} =Area

Multiply the top and bottom:

5*7 = 35

9*8 = 72

35/72 = Area

6 0
3 years ago
Suppose that 500 parts are tested in manufacturing and 10 are rejected.
alexdok [17]

Answer:

a) z=\frac{0.02 -0.03}{\sqrt{\frac{0.03(1-0.03)}{500}}}=-1.31  

p_v =P(Z  

If we compare the p value obtained and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of rejected items is less than 0.03.  

b) We can use a 95 percent upper confidence interval.

On this case we want a interval on this form : (-\infty,\hat p +z_{\alpha}\sqrt{\frac{\hat p (1-\hat p)}{n}})

So the critical value would be on this case Z_{\alpha}=1.64 and we can use the following excel code to find it: "=NORM.INV(1-0.05,0,1)"

We found the upper limit like this:

0.02+1.64\sqrt{\frac{0.02 (1-0.02)}{500}}=0.03026

And the interval would be: (-\infty,0.03026)

And since our value (0.02) is contained in the interval We fail to reject the hypothesis that p=0.03

Step-by-step explanation:

Part a

Data given and notation  

n=500 represent the random sample taken

X=10 represent the number of objects rejected

\hat p=\frac{10}{500}=0.02 estimated proportion of objects rejected

p_o=0.03 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 70% of adults say that it is morally wrong to not report all income on tax returns.:  

Null hypothesis:p=0.03  

Alternative hypothesis:p < 0.03  

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{0.02 -0.03}{\sqrt{\frac{0.03(1-0.03)}{500}}}=-1.31  

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 one tailed left test the p value would be:  

p_v =P(Z  

If we compare the p value obtained and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of rejected items is less than 0.03.  

Part b

We can use a 95 percent upper confidence interval.

On this case we want a interval on this form : (-\infty,\hat p +z_{\alpha}\sqrt{\frac{\hat p (1-\hat p)}{n}})

So the critical value would be on this case Z_{\alpha}=1.64 and we can use the following excel code to find it: "=NORM.INV(1-0.05,0,1)"

We found the upper limit like this:

0.02+1.64\sqrt{\frac{0.02 (1-0.02)}{500}}=0.03026

And the interval would be: (-\infty,0.03026)

And since our value (0.02) is contained in the interval We fail to reject the hypothesis that p=0.03

3 0
3 years ago
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