11.4 multiplied by 4 trucks = 45.6 tons in the trucks.
55 - 45.6 = what remains in the pile which would be 9.4 tons.
Answer:
No; the absolute value of a negative number is equal to its opposite
Step-by-step explanation:
Answer:
$8659
Step-by-step explanation:
Patricia's sales = $ 72,900
Her commission = 11% of $72,900
= 11/100 * 72,900 [ % means out of 100 and of means to multiply ]
= 11 * 729
= $ 8019
Her monthly salary = $640
Her monthly salary along with commission = $640 + $8019
= $8659
![\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>
~AH1807</h3>
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
