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insens350 [35]
3 years ago
15

What is 27/40 as a decimal? Explain how you got your answer.

Mathematics
2 answers:
Sedbober [7]3 years ago
8 0

Answer:

0.675

Step-by-step explanation:

27 ÷ 40

Ipatiy [6.2K]3 years ago
5 0
To write 27/40 as a decimal you have to divide numerator by the denominator of the fraction.
We divide now 27 by 40 what we write down as 27/40 and we get 0.675
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Answer:

C y=800(1.15)^12

Step-by-step explanation:

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3 years ago
4/5 divided by 9/10 as a fraction
kirill [66]

Answer:

8/9

Step-by-step explanation:

(4/5)/(9/10)=(4*10)/(5*9)

(4*10)/(5*9)= 40/45

40/45= 8/9

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3 years ago
Anita is making a curtain to surround a table. She bought 20 yards of fabric. Her total cost was $13. What was the cost per yard
Ne4ueva [31]

In other words, what's the unit rate, the price per 1 yard of fabric?

$13

--------- = ($13 / 20 yd), or $0.65 / yd

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3 years ago
Read 2 more answers
Select all numbers that are sulutions to the inequality w>5
sertanlavr [38]
<h3>W is any number greater than 5</h3>

<em>Because there is a sign of greater than between w and 5</em>

So, By this info we can say that; Option A, Option D, Option E are correct!

8 0
3 years ago
Read 2 more answers
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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