Trying to Make Money - What Should I Put on Fliers?

You might be better off going door to door and asking if anyone needs any help. You could print up business cards with your name and number and hand them out when you go talk to people

1. What is going on with my kitten?

Strange, I never had a kitten do that it MAY be affection because of the licks and purrs I did however have a kitten that would bite purr then lick but he was playing but you said that she comes up straight to your face and your armpits...... She MAY just be trying to play not understanding that your pushing her away and not trying to play try closing you bedroom door a couple of nights(Like maybe 2 weeks) so she can not get in and bite you

2. How can i improve the average coefficient of performance household refrigerator? thanks?

Put a one way mirror in door and a light switch so when you turn it on you can see whats inside without opening the door. Low tech but maybe it works

3. Do you think it's rude?

I remember as a child going to one of my friends birthday parties every year on the invite it said please no gifts. One year it said please bring a canned food item for the food shelf and then for her birthday we divided up into two teams and went door to door collecting food for the local food pantry. It was a game, certain types of food were worth different amounts of points. I remember sardines were worth the most and one lady gave us like 6 cans and we won on that alone. It was so much fun and it was a charitable thing too. People want to give gifts for birthdays, there is nothing wrong with asking them not to and giving them a suggestion of what they could do instead with their money. I hope it goes well and Happy birthday to your daughter!

4. How to open car door with door panel and lock removed?

The door latch release is inside the door close to the lock hole. Reach in there with a pick and open it ffs

5. Garage door question. What is the rubber piece called on the bottom of door that is threaded into door. ?

On a garage door it's called an astragal, but it's not "threaded" to the door. On a steel door they slide down a retainer. On wood doors they are usually nailed w/ roofing nails.

6. How can I rehang heavy fire doors?

In your specific case, the weight of the door has probably stressed the screws in the jambs and enlarged the jamb holes a bit, making it difficult to tork the hinges snug, thus causing the door to sag. Since you already have a good shim between the jamb and stud, replace the short screws with some 2 to 2 1/2 inch wood screws. Typically size 10 to 12. Now, all the weight will be supported by the stud instead of the worn jamb. Also inspect the screws in the door side of the hinge to be sure they are secure. Longer screws there may help it bit if there is enough solid wood to bite to. If they are a bit loose also, put some wood shims coated with wood glue in the screw holes, let it cure and reinstall the hinges. I should have mentioned, be sure to check the plumb of the jambs and adjust as required before reattaching the hinges and door

7. Monty Hall Revisited: Winning Both Goats!

but I thought I would give a closer look at how these relate in general.Say we have two events that we are interested in, call them $A$ and $B$. We wonder what the probability is that both will occur, that is, we want to know $mathbb P(A cap B)$ where $cap$ denotes the intersection of the two events. We can always decompose a joint probability (a probability of multiple things) into conditional probabilities: $mathbb P(A cap B) = mathbb P(A) mathbb P(B|A)$ where $mathbb P(B|A)$ is the conditional probability that $B$ occurs, given that we already know that $A$ occurred (note that the other decomposition, $mathbb P(A cap B) = mathbb P(B) mathbb P(A|B)$, is equivalent; the two individual terms may differ, but their product will be the same).If the two events are independent (i.e. the outcome of one has no influence on the outcome of the other), then $mathbb P(B|A) = P(B)$ because the probability distribution for $B$ wo not change regardless of the value for $A$ that we observe.Now, on to the problem at hand. Suppose that we've chosen door 1 and are deciding whether we should switch (the argument holds for any door we choose, just relabel the door names). We want to know what the probability is that there are goats behind doors 2 and 3; call these events $g_2$ and $g_3$, and we want $mathbb P(g_2 cap g_3)$. We know that this is the same as $mathbb P(g_2) mathbb P(g_3|g_2)$. There are two goats and one car, so if they are distributed uniformly at random, there's a 2/3 chance of a goat behind each door, so $mathbb P(g_2) = 2/3$. Note that $mathbb P(g_3) = 2/3$ too, but we are interested in $mathbb P(g_3|g_2)$. If we know that there is a goat behind door two, there's only one goat left (and one car), so $mathbb P(g_3|g_2)=1/2$; clearly $g_3$ and $g_2$ are not independent; this was the error that Monty made. Thus we get $$mathbb P(g_2 cap g_3) = mathbb P(g_2) mathbb P(g_3|g_2) = frac23 * frac12 = frac13

eq mathbb P(g_2) mathbb P(g_3) = frac23 * frac23 = frac49$$ Since the probability of both other doors having goats is 1/3, the probability of at least one of them having a car is 2/3, as intuited. Note that this is a common method in probability - if you want to know the probability that at least one of a set of events occurs, it's often easiest to determine this as $mathbb P(ge 1 text event occurs) = 1 - mathbb P(textno event occurs)$.