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NARA [144]
3 years ago
7

The ratio of the number of boys to the number of girls on the swim team is 4:2. If there are 24 athletes on the swim team, how m

any boys than girls are there?
Mathematics
1 answer:
Daniel [21]3 years ago
6 0

Answer:

We have 8 more boys than girls on the swim team.

Step-by-step explanation:

The ratio of boys to girls on a swim team is represented as:

4 : 2

Total number of athletes is given as 24

Total ratio is given as:

4 + 2 = 6

Step 1

We find the number of boys and girls

Number of boys = 4/6 × 24 = 16 boys.

Number of girls = 2/6 × 24 = 8 girls.

Therefore, since we have 16 boys and 8 girls on the swim team, we have 8 more boys than girls on the swim team.

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x = 109°

3*180° - 100° - 102° - 108° - 121° = 109°

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At the beginning of a snowstorm, Grayson had 10 inches of snow on his lawn. The snow then began to fall at a constant rate of 0.
maxonik [38]

Answer:

  • Grayson had 10 inches of snow in his lawn
  • the snow falls at a constant rate of 0.5 inches per hour
  • no snow was melting
  • the snow has fallen for 3 hours  

Let y be the height of the snow in the lawn

Grayson had 10 inches at the beginning (at t=0) so y= 10

3 hours passed so y= 10 + 3*0.5 = 11.5 inches

Grayson had 11.5 inches after three hours

Assuming that the snow continued to fall for an unkhown time t

y = 0.5*t + 10

we had 10 at the beginning so we add 10 (+10)

the snow is growing by 0.5 inches in the single hour so we multiply by 0.5

t is the time and it can 4, 5, 6 ...... hours

Just like a function :

  • 0.5 is the slope
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3 0
3 years ago
Josh has a box of ballons.in that box he finds that 1/8 of the ballons are blue out of 120 ballons in the box the rest is red.ho
Stolb23 [73]

Answer:

15

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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.

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3 years ago
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