It is 28 because you have to take the median out then go to the first set of data and find the median of that.
Answer:
5%
Step-by-step explanation:
Hospital A (with 50 births a day), because the more births you see, the closer the proportions will be to 0.5.
Hospital B (with 10 births a day), because with fewer births there will be less variability.
The two hospitals are equally likely to record such an event, because the probability of a boy does not depend on the number of births
Answer:
1296 cm²
Step-by-step explanation:
36² = 1296 cm²
<em>H</em><em>O</em><em>P</em><em>E</em><em> </em><em>T</em><em>H</em><em>I</em><em>S</em><em> </em><em>H</em><em>E</em><em>L</em><em>P</em><em>S</em><em> </em><em>A</em><em>N</em><em>D</em><em> </em><em>H</em><em>A</em><em>V</em><em>E</em><em> </em><em>A</em><em> </em><em>N</em><em>I</em><em>C</em><em>E</em><em> </em><em>D</em><em>A</em><em>Y</em><em> </em><em><</em><em>3</em>
<h3>
Answer:</h3>
- <u>20</u> kg of 20%
- <u>80</u> kg of 60%
<h3>
Step-by-step explanation:</h3>
I like to use a little X diagram to work mixture problems like this. The constituent concentrations are on the left; the desired mix is in the middle, and the right legs of the X show the differences along the diagonal. These are the ratio numbers for the constituents. Reducing the ratio 32:8 gives 4:1, which totals 5 "ratio units". We need a total of 100 kg of alloy, so each "ratio unit" stands for 100 kg/5 = 20 kg of constituent.
That is, we need 80 kg of 60% alloy and 20 kg of 20% alloy for the product.
_____
<em>Using an equation</em>
If you want to write an equation for the amount of contributing alloy, it works best to let a variable represent the quantity of the highest-concentration contributor, the 60% alloy. Using x for the quantity of that (in kg), the amount of copper in the final alloy is ...
... 0.60x + 0.20(100 -x) = 0.52·100
... 0.40x = 32 . . . . . . . . . . .collect terms, subtract 20
... x = 32/0.40 = 80 . . . . . kg of 60% alloy
... (100 -80) = 20 . . . . . . . .kg of 20% alloy