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natima [27]
3 years ago
14

How do I know if two ratios are equal to make a proportion

Mathematics
1 answer:
AfilCa [17]3 years ago
3 0
One way to see if two ratios are proportional is to write them as fractions and then reduce them. If the reduced fractions are the same, your ratios are proportional. 
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-2/3 minus (-4/6)= <br>​
lina2011 [118]

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-  \frac{2}{3}  - ( -  \frac{4}{6} ) =  \\

-  \frac{2}{3}  +  \frac{4}{6}  =  \\

-  \frac{2}{3}  +  \frac{2 \times 2}{2 \times 3}  =  \\

-  \frac{2}{3}   +  \frac{2}{3}  =  0\\

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8 0
3 years ago
3a + 10 − 4q + 7a − 4q − 6.
soldi70 [24.7K]

Answer:

Step-by-step explanation:

3a+10-4q+7a-4q-6  combine terms

10a-8q+4

7 0
2 years ago
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A random sample of 25 ACME employees showed the average number of vacation days taken during the year is 18.3 days with a standa
Norma-Jean [14]

Answer:

a) Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

b) df=n-1=25-1=24  

For this case the p value is given p_v = 0.0392

If we compare the p value and 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, so we can conclude that the true mean is higher than 15 at 5% of signficance.  

c) Type I error, also known as a “false positive” is the error of rejecting a null  hypothesis when it is actually true. Can be interpreted as the error of no reject an  alternative hypothesis when the results can be  attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

Step-by-step explanation:

Data given and notation  

\bar X=18.3 represent the sample mean

s=3.72 represent the sample standard deviation

n=25 sample size  

\mu_o =15 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

Part a: State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean for vacation days is higher than 15, the system of hypothesis would be:  

Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Part b: P-value  and conclusion

The first step is calculate the degrees of freedom, on this case:  

df=n-1=25-1=24  

For this case the p value is given p_v = 0.0392

Conclusion  

If we compare the p value and 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, so we can conclude that the true mean is higher than 15 at 5% of signficance.  

Part c

Type I error, also known as a “false positive” is the error of rejecting a null  hypothesis when it is actually true. Can be interpreted as the error of no reject an  alternative hypothesis when the results can be  attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

3 0
3 years ago
An athlete training for the olympics swims each lap of a four-lap race in the following times: 54.73 , 54.56 , 54.32 , and 54.54
solniwko [45]

Answer:

218.25 seconds

Step-by-step explanation:

54.73

54.56

54.32

54.54

+_____

218.15

3 0
3 years ago
Can you please solve this please and thank you.
Lilit [14]

Answer:

The answer is 3

Step-by-step explanation:

It’s 3 because with dilation your multiplying points to make a shape bigger, so multiply all the numbers by 3 and when you do that you woulda gotten what’s shown in B

4 0
2 years ago
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