You need 300 ml of a 75% alcohol solution. on hand, you have a 10% alcohol mixture and a 85% alcohol mixture. how much of each mixture will you need to obtain the desired solution?

Well, to find how much of the Alcohol Solution you will need, you will need to do;

300 (divided by) 10, which will give you a answer of 30. Then you will do 30 (times) by 7, which will give you 210, which will then give you 70% of your 75%. Then, to find the other 5%, you need to do what the answer was to 10%, and divide it by 2, so in this case, the answer will be 15. So now to get your full 75%, you add your working out together for the 70% and the 5% which gives you 225.

Alcohol Solution = 225

Then, to find out what amount of the Alcohol Mixture you need, you will do;

300 (divided by) 10 which is 30, which will give you the answer to how much Alcohol Mixture you need.

Alcohol Mixture {1} = 10

To find out the second amount of how much of the other Alcohol Solution you will need, you need to;

Find 10% of 300, which is 30 (300 {divided by} 10), then you will times that answer (30), by 80 to find your 80%, which will be 240, then you will need to find your other 5%, so you will need to do 300 (divided by) half of the answer you got before you timed the answer by 8, so that would be 15 (added onto) 240, which will give you an overall answer of 255.

Alcohol Mixture {2} = 255

The ones in bold are your answers in order, hope they help! xo

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.