11.7
Formula: a^2 + b^2 = c^2
so substitute the numbers like this:
10^2 + 6^2 = c^2
both 10^2 and 6^2 are perfect squares:
10 × 10= 100 and 6 × 6= 36
100 + 36= c^2
add:
100 + 36 =
136 = c^2
now in order to get rid of the squared, you put it into a radical:
√136 = √c^2
note: whatever you do on one side, you do to the other
it's easier to use the calculator for this so put √136 and you're gonna get 11.66
since it's asking you to round it to the nearest tenth:
11.66 = 11.7
sorry if the explanation is kinda long ✋ i tried explaining it better in the end but i failed
Answer:
a)
b) ![P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668](https://tex.z-dn.net/?f=P%28X%3E%202%29%3D1-P%28X%5Cleq%202%29%3D1-%5B0.0211%2B0.0995%2B0.211%5D%3D0.668)
c)
Step-by-step explanation:
1) Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
2) Solution to the problem
Let X the random variable of interest, on this case we now that:
The probability mass function for the Binomial distribution is given as:
Where (nCx) means combinatory and it's given by this formula:
Part a
Part b
![P(X> 2)=1-P(X\leq 2)=1-[P(X=0)+P(X=1)+P(X=2)]](https://tex.z-dn.net/?f=P%28X%3E%202%29%3D1-P%28X%5Cleq%202%29%3D1-%5BP%28X%3D0%29%2BP%28X%3D1%29%2BP%28X%3D2%29%5D)
![P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668](https://tex.z-dn.net/?f=P%28X%3E%202%29%3D1-P%28X%5Cleq%202%29%3D1-%5B0.0211%2B0.0995%2B0.211%5D%3D0.668)
Part c
Answer:
-1/3, -1/4, -3/10, -1/8, -1/16, 1/3, 1/4, 3/10, 1/8, 2/5.
Step-by-step explanation:
-2/5 is -0.4 and 1/2 is 0.5
Ten rational numbers in between are, -0.33333 (-1/3), -0.25 (-1/4), -0.3 (-3/10), -0.125 (-1/8), -0.0625 (-1/16), 0.33333 (1/3), 0.25 (1/4), 0.3 (3/10), 0.125 (1/8), 0.4 (2/5)