Answer:
1.42 hours
Step-by-step explanation:
In this case we must take into account that the speed would be to fill a pool for a time, that this case is in hours, therefore it would be like this:
1/5 for the first pipe
1/2 for the second pipe
Now, we will use both at the same time, therefore:
1/5 + 1/2 = 1 / x
0.7 = 1 / x
x = 1 / 0.7
x = 1.42
That is to say that if we use both we will take a total of 1.42 hours
The answer is <span>mean = 13,027; median = 12,200; no mode
</span>
Let's rearrange values from the lowest to the highest:
11350, 12050, 12200, 13325, 16211
<span>The mean is the sum of all values divided by the number of values:
</span>(11350 + 12050 + 12200 + 13325 + 16211)/5 ≈ 13027
The median is the middle value. If there is an odd number of data, then the median is the value in the middle. In the data set 11350, 12050, 12200, 13325, 16211, the median (the middle value) is 12200
<span>The mode is the value that occurs most frequently. Since none of the number does not occur most frequently, there is no mode.
</span>
19z-7 - ( 5z-9 + <span>8z+3)
= 19z - 7 - (13z - 6)
= 19z - 7 - 13z + 6
= 6z - 1
answer
remaining side = </span> 6z - 1
<em>u={1,2,3,4,5},A={2,4} and Beta {2,5,5}</em>
<em>now</em><em>,</em><em> </em><em>(AUB)</em><em>=</em><em>{</em><em>1</em><em>,</em><em>3</em><em>,</em><em>3</em><em>,</em><em>4</em><em>,</em><em>5</em><em>}</em>
<em>[</em><em>AUB</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>set</em><em> </em><em>of</em><em> </em><em>all</em><em> </em><em>elements</em><em> </em><em>of</em><em> </em><em>set</em><em> </em><em>A</em><em> </em><em>and</em><em> </em><em>set</em><em> </em><em>B</em><em> </em><em>without </em><em>any</em><em> </em><em>repetition </em><em>]</em>
<em>n</em><em>(</em><em>AUB</em><em>)</em><em>=</em><em>5</em>
<em>n</em><em>(</em><em>AUB</em><em>)</em><em>is</em><em> </em><em>the</em><em> </em><em>total</em><em> </em><em>no</em><em> </em><em>of</em><em> </em><em>elements</em><em> </em><em>in</em><em> </em><em>set</em><em> </em><em>(</em><em>AUB</em><em>)</em>