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ANEK [815]
4 years ago
9

A piano has white keys and black keys in the ratio of 13 : 9. If there are 52 white keys,how many black keys are there?

Mathematics
2 answers:
Hitman42 [59]4 years ago
5 0
✡ Answer: 9/36 <span>✡
</span><span>
- - Why?

Because if you look at it like this:

</span>\frac{13}{52} =  \frac{9}{36}
<span>
- 52 being the white and 36 because the black.

</span>✡Hope this helps<span>✡</span><span>



</span>
luda_lava [24]4 years ago
3 0
Let w = white and b = black.
Using the proportion,
13/w = 9/b
Since w = 52 white keys, therefore
13/52 = 9/b
b = 36 black keys
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seraphim [82]

Answer:

z=\frac{0.733 -0.5}{\sqrt{\frac{0.5(1-0.5)}{30}}}=2.552  

p_v =2*P(z>2.552)=0.0107  

So the p value obtained was a very low value and using the significance level assumed \alpha=0.05 we have 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 songs selectd at random  differs from 0.5 .  

There is strong evidence that the proportion of songs downloaded by Ted and Suzanne differs from 0.5.

Step-by-step explanation:

Data given and notation

n=30 represent the random sample taken

X=22 represent the number of songs selectd at random

\hat p=\frac{22}{30}=0.733 estimated proportion of songs selectd at random

p_o=0.5 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 the true proportion of random songs selected is 0.5 (random) or no.:  

Null hypothesis:p=0.5  

Alternative hypothesis:p \neq 0.5  

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{0.733 -0.5}{\sqrt{\frac{0.5(1-0.5)}{30}}}=2.552  

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 assumed for this case is \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

p_v =2*P(z>2.552)=0.0107  

So the p value obtained was a very low value and using the significance level assumed \alpha=0.05 we have 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 songs selectd at random  differs from 0.5 .  

There is strong evidence that the proportion of songs downloaded by Ted and Suzanne differs from 0.5.

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Step-by-step explanation:

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