Answer:
f(x) = 34,500 × (1.032)^x
Step-by-step explanation:
Just completed the test and copied the answer directly from the review.
Prime factor is a factor that is a prime number.
Prime factors of :
8 : 2 × 2 × 2. Also can be written as
12 : 2 × 2 × 3. Also can be written as
20 : 2 × 2 × 5. Also can be written as
30 : 2 × 3 × 5
56 : 2 × 2 × 2 ×7. Also can be written as
70 : 2 × 5 × 7
You can find all of them by using Factor Tree [ as shown in the picture] or Dividing by prime numbers.
I'm not sure about the second image.
But the first one is telling you to translate x, 3 units left, and translate y, 4 units up.
(-6, 0) will turn to (-9, 4)
I subtracted 3 from -6 and added 4 to 0.
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
answer is A, if its not A then B
