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cluponka [151]
1 year ago
8

Eddie's fishing. He has 1 1/4 ounces of weights on his line, but his bait isn't getting to the bottom of the lake. Eddie add ano

ther 1/2-ounces weight to the line, but it's still not enough. Finally, Eddie adds a 3/4-ounces weights and the bait sinks. How much weight does eddie now have on his line?
Mathematics
2 answers:
mario62 [17]1 year ago
8 0
The initial weight is given by:
 1  \frac{1}{4}  = 1.25

 Then, we add a second weight:
 \frac{1}{2}  = 0.5

 Finally, we add a third weight:
 \frac{3}{4}  = 0.75

 Adding the three added weights, we can find the total weight.
 We have then:
 1.25 + 0.5 + 0.75 = 2.5

 The total weight is:
 2.5 = 2  \frac{1}{2} 

 Answer:
 
eddie now has 2  \frac{1}{2} ounces on his line
mel-nik [20]1 year ago
8 0

Answer:

idk

Step-by-step explanation:

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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

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Part d

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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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The shelf should sell for $235.20.

Marking the price up by 60% means taking 160% of the cost:
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Answer:

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A geometric setting does not have a set number of trials, and the variable in question is the number of trials it takes to get the first success.

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