Answer:
Building A is 660 feet and Building B is 830 feet
Step-by-step explanation:
Let x represent the height of building B.
Since building A is 170 feet shorter than building B, it can be represented by x - 170.
Create an equation and solve for x:
(x) + (x - 170) = 1490
2x - 170 = 1490
2x = 1660
x = 830
So, the height of building B is 830 feet.
Subtract 170 from this to find the height of building A:
830 - 170
= 660
Building A is 660 feet and Building B is 830 feet
We have five value in the data-set
The third value will be 10 since we want the median to be 10
We want the mean to be 14
To find the mean of a data set, we divide the sum of the values by the number of values
Mean = Sum of values ÷ Number of values
14 = Sum of values ÷ 5
Sum of values = 14 × 5
Sum of values = 70
So we need 5 values that add up to 70, one of the value is 10, which is the median. We would want two values that are smaller than 10 and two values more than 10.
These four value must add up to 70 - 10 = 60
From here we can do trial and error:
Choose any two values less than 10, say 9 and 8
We now have in total 8 + 9 + 10 = 27
We have 70 - 27 = 43 left to find
Choose any two values that are bigger than 10 that add up to 43, for example, 20 and 23
Now we have our 5 values;
8 9 10 20 23
Do the checking bit:
We can see from the set, the median is 10
Mean = [8+9+10+20+23] ÷ 5 = 70 ÷ 5 = 14
We can have values other than 8, 9, 20 and 23 as long as two values smaller than 10 and two values more than 10. All five values must add up to 70.
<span>the answer is 2(19d+17)</span>
Answer:
0.4
Step-by-step explanation:
so, Katherine can do 0.1 of the job per hour. (if she needs 10 hours for the job, each hour she can do the whole divided by the time, so 1/10, which is 0.1)
if in two hours she and Marina can be done, then Marina will do 1(the whole job)-2/10 (twice as much as she can do in an hour)=0.8 of the job. So if Marina does 0.8 of the job in two hours, each hour she will do half of it: 0.4 - and this is the correct answer
The answer is D one trillion percent
