175 mL at 25% concentration of alcohol contains 0.25 (175 mL) = 43.75 mL of alcohol. If <em>v</em> is the amount of the 70% solution that you use, then that amount contains 0.7<em>v</em> mL of alcohol.
Mixing these two yields a total volume of 175 mL + <em>v</em>, and it contains 43.75 mL + 0.7<em>v</em> alcohol. You want to end up with a concentration of 45%, which means the ratio of the amount of alcohol to the total volume needs to be 0.45:
(43.75 mL + 0.7<em>v</em>) / (175 mL + <em>v</em>) = 0.45
Solve for <em>v</em> :
43.75 mL + 0.7<em>v</em> = 0.45 (175 mL + <em>v</em>)
43.75 mL + 0.7<em>v</em> = 78.75 mL + 0.45<em>v</em>
0.25<em>v</em> = 35 mL
<em>v</em> = 140 mL
by pythagorean formula, the last side is √(61)
by cos rule
cos A
A = 39.81
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
Answer:
Assuming one unit is one unit of measurement the answer would be nine
Step-by-step explanation:
