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Vilka [71]
1 year ago
5

The function y=7500(1.08)t represents the value y of an investment after t years. What is the value of the investment after 6 ye

ars?
Mathematics
1 answer:
boyakko [2]1 year ago
4 0

Answer:

The answer is 48600, but I'm not entirely sure if it correct.

From what I know, if I were to see ¨y=7500(1.08)t¨, we could have to substitute the t with the number of years. Looking like this, ¨y=7500(1.08) x 6¨.

The reason I don know if my answer is correct or not since I don know if you have the t next to 1.08 to represent that or to simply multiply the property of it with 6 (the number of years).

So I tried...

Step-by-step explanation:


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Equivalent equations are equations that have exactly the same solution. To find equivalent equations, it is best to solve them individually to find the solutions. The ones with the same solution in the end are considered equivalent equations.


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At Central High School, there are 48 football players and 20 cheerleaders. What is the ratio of cheerleaders to football players
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5 : 12

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According to a Los Angeles Times study of more than 1 million medical dispatches from 2007 to 2012, the 911 response time for me
Reil [10]

Answer:

a) \bar X=10.65

Median =\frac{10.7+10.7}{2}=10.7

Mode= 10.7

b) Range = 11.8-8.3=3.5

s= 0.948

c)IQR = Q_3 -Q_1 = 11.05-10.55=0.5

And we can find the usual limits with:

Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8

Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

d) The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

Step-by-step explanation:

We have the following data:

11.8 10.3 10.7 10.6 11.5 8.3 10.5 10.9 10.7 11.2

Part a

We can calculate the sample mean with the following formula:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

And if we replace we got: \bar X=10.65

For the median we need to sort the values on increasing order and we have:

8.3 10.3 10.5 10.6 10.7 10.7 10.9 11.2 11.5 11.8

Since n =10 we can calculate the median as the average between the 5th and 6th position of the dataset ordered and we got:

Median =\frac{10.7+10.7}{2}=10.7

The mode would be the most repeated value on this case:

Mode= 10.7

Part b

The range is defined as Range =Max-Min and if we replace we got:

Range = 11.8-8.3=3.5

We can calculate the standard deviation with the following formula:

s= sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And if we replace we got:

s= 0.948

Part c

For this case we can use the IQR method in order to determine if 8.3 is an outlier or not.

We can calculate the first quartile with these values: 8.3 10.3 10.5 10.6 10.7 10.7 and Q_1= \frac{10.6+10.7}{2}=10.55

And for the Q3 we can use: 10.7 10.7 10.9 11.2 11.5 11.8 and we got Q_3 = \frac{10.9+11.2}{2}=11.05

Then we can find the IQR like this:

IQR = Q_3 -Q_1 = 11.05-10.55=0.5

And we can find the usual limits with:

Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8

Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

Part d

The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

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2 years ago
Day care centers expose children to a wider variety of germs than the children would be exposed to if they stayed at home more o
USPshnik [31]

Answer:

a) This is an Observational Study because in this kind of study investigators observe subjects and measure variables of interest without assigning treatments to the subjects. Here, the Gilham et al. (2005) studied two different groups where no treatment or intervention was done. These groups were independent of each other.

b) proportions of children with significant social activity in children with acute lymphoblastic leukemia = 1020/1272 = 0.80

proportions of children with significant social activity in children without acute lymphoblastic leukemia = 5343/6238 = 0.86

c) Odds ratio can be calculated using the following formula:

OR= \frac{a/b}{c/d}

where: a - Number in exposed group with positive outcome(here this means number of children with significant social activity associated with acute lymphoblastic leukemia)

b- Number of children without social activity having with acute lymphoblastic leukemia

c- Number of children with social activity having without acute lymphoblastic leukemia

d- Number of children without social activity having without acute lymphoblastic leukemia

OR= \frac{1020/252}{5343/895}

OR= 0.6780

d) The 95% confidence interval of this Odds Ratio is 0.5807 to 0.7917.

e) Since the odds ratio lies in this confidence interval indicate that the amount of social activity is associated with acute lymphoblastic leukemia. The children with more social activity have a higher occurrence of acute lymphoblastic leukemia.

8 0
1 year ago
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