Answer:
Step-by-step explanation:
We can factor the expression by using the common factor.
Pull out 3x from the expression.
Common Factor is similar to dividing by the term we pull out. When we pull out the term, we divide the whole expression and keep the term out of the expression.
Answer:
$409 per month
Step-by-step explanation:
If josefina paid $2045 over a time period of 5 months, then we have to divide how much she paid ($2045) by 5 to find out how much she paid for rent in 1 month.
We can use the equation: 2045 ÷ 5
When we solve that equation, we get 409 which means that josefina paid $409 per month for rent.
(sorry if this is confusing but I hope it helped)
The area is 192 and the speed it’s increasing at is 24 cm^2/s
Yes it does, let me try to answer it!
Answer:
1) you're going to have to flip the coins (or fake numbers) for the experimental trials.
2) for the theoretical, there is 1/2 chance for heads or tails with each toss, so you'd expect that out of 10 tosses, 5 heads, 5 tails. out of 100 tosses- 50 heads, 50 tails.
When tossing 2 coins- 1/2×1/2 = 1/4 (25%) chance that 2 heads, 2 tails, or 1 heads & 1 tails. Deviation value comes from after you done your flipping and recorded your data. So if on 100 flips you actually got 50 and 50 (rarely us that exact ;), the deviation from the expected of 50/50 would be 0.00. If however you flipped 100 heads or 100 tails (impossible), then the deviation value would be 1.00.
|(100-50)| ÷ 50 = 50÷50 = 1.00
So usually you may have data like: 47/53 or something a little off than 50/50, making deviation |(47-50)| ÷ 50 = 3÷50 = 0.06.
Now the number of flips is important for the outcome! So if a coin toss if 10 times had 4 heads, 6 tails, the deviation value would be:
|(4-5)| ÷ 5 = 1÷5 = 0.20
So increasing the # flips DECREASES the deviation value!!
Whether it's from 10 to 100, or from 100 to 200. Look at my example of how the 10-flip deviation of 0.20 decreased to 0.06 with 100-flip
