Answer:
43.6 (approximately)
Step-by-step explanation:
Diagonal measures of the screen is,
√(35²+26²)
= √(1901)
= 43.6 (approximately)
Answered by GAUTHMATH
Answer:
0.0416
Step-by-step explanation:
Given :
Sample size, n = 300
Sample mean, x = 35.5
Population mean, m = 35
Standard deviation, s = 5
The test statistic :
Zstatistic = (x - m) / s/sqrt(n)
Zstatistic = (35.5 - 35) / 5/sqrt(300)
Zstatistic = 0.5 / 0.2886751
Zstatistic = 1.732
Using the p value calculator from Zstatistic :
One tailed P value at 95% confidence interval is : 0.0416
<span>(–7 + 3i) – (2 – 6i)
subtract the like terms.
-7 - 2 = -9
3i - (-6i) = 3i +6i = 9i
The answer becomes -9 + 9i</span>
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:
200000x
Step-by-step explanation:
