Given :
Rema purchased 18 ounces of gold in 2005 for $414 per ounce in order to try to diversify her investment portfolio.
She sold a third of her holdings in gold in 2009 at a price $953 per ounce. She sold the rest of her gold holdings in 2010 for $1,367 per ounce.
To Find :
What is Prema’s profit in 2009 and 2010.
Solution :
Price of one third ounces of gold, in 2005 is :
P = $414/3 = $138 .
Profit from selling them in 2009 is :
P1 = $(953 - 138) = $815 .
Profit from selling them in 2010 is :
P2 = $( 1367 - ( 414 - 138 ) ) = $1091 .
Therefore, Prema's profit in 2009 is $815 and 2010 is $1091.
Hence, this is the required solution.
9514 1404 393
Answer:
D. $101,000 – $120,000
Step-by-step explanation:
The bar graph is not completely labeled, but in the context of the question it seems safe to assume that the vertical scale can be considered to represent relative frequency.
So, the shortest bar is the one with the lowest frequency. The horizontal scale identifies that as 101-120. If we assume that is salary in thousands of dollars, then Choice D is appropriate.
Answer:
Step-by-step explanation:
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Answer:
123.1 pounds is vary long, and I don't want to repeat, so 55.8372207 repeat.
Step-by-step explanation:
If you have any questions regarding my answer, tell me them in the comments, and I will come answer them for you. Have a good day.
Answer:
1. D. 20, 30, and 50
2. A. 86
3. B. 94
Step-by-step explanation:
1. To find the outliers of the data set, we need to determine the Q1, Q3, and IQR.
The Q1 is the middle data in the lower part (first 10 data values) of the data set (while the Q3 is the middle data of the upper part (the last 10 data values) the data set.
Since it is an even data set, therefore, we would look for the average of the 2 middle values in each half of the data set.
Thus:
Q1 = (85 + 87)/2 = 86
Q3 = (93 + 95)/2 = 94
IQR = Q3 - Q1 = 94 - 86
IQR = 8
Outliers in the data set are data values below the lower limit or above the upper limit.
Let's find the lower and upper limit.
Lower limit = Q1 - 1.5(IQR) = 86 - 1.5(8) = 74
The data values below the lower limit (74) are 20, 30, and 50
Let's see if we have any data value above the upper limit.
Upper limit = Q3 + 1.5(IQR) = 94 + 1.5(8) = 106
No data value is above 106.
Therefore, the only outliers of the data set are:
D. 20, 30, and 50
2. See explanation on how to we found the Q1 of the given data set as explained earlier in question 1 above.
Thus:
Q1 = (85 + 87)/2 = 86
3. Q3 = (93 + 95)/2 = 94
