when a large object fell into a "full" swimming pool, it displaced 35.1 gallons of water. The density of the object is 1.65 g/mL. What is the object's weight in pounds?

m = 483.32 pounds

Explanation:

Given that,

The density of the object is 1.65 g/mL

Volume of the water is 35.1 gallons

We know that,

1 gallon = 3785.41 mL

35.1 gallons = 132868 mL

The density of an object is equal to its mass divided by its volume. So,

Also, 1 pound = 453.592 grams

So, 219232.2 grams = 483.32 pounds

So, the weight of the object is 483.32 pounds.

linda: 3 ; laura: 15

step-by-step explanation:

note that linda + laura's age is 18, and that linda is 12 years younger than her sister.

let "linda" = x , "laura" = y:

x + y = 18

x = y - 12

first, plug in y - 12 for x in the first equation:

(y - 12) + y = 18

simplify.

2y - 12 = 18

isolate the variable (y). note the equal sign, what you do to one side, you do to the other. do the opposite of pemdas. first add, then divide.

2y - 12 (+12) = 18 (+12)

2y = 18 + 12

2y = 30

(2y)/2 = (30)/2

y = 30/2

y = 15

next, plug in 15 for y in the second equation (the other equation would work too).

x = y - 12

x = (15) - 12

x = 3

linda's age is 3, laura's age is 15.

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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.