<span>A right triangle has a 28° angle.
As you know a</span> right triangle<span> is a </span>triangle<span> in which one angle is a </span>right angle that equal 90° angle.
Now you know 2 angles values: 28° and 90°
So the third angle = 90° - 28° = 62°
Answer:
28°, 62°, 90°
Answer:
In the long run, ou expect to lose $4 per game
Step-by-step explanation:
Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n^2 if heads comes up first on the nth toss.
Assuming X be the toss on which the first head appears.
then the geometric distribution of X is:
X 
geom(p = 1/2)
the probability function P can be computed as:
where
n = 1,2,3 ...
If I agree to pay you $n^2 if heads comes up first on the nth toss.
this implies that , you need to be paid 
![\sum \limits ^{n}_{i=1} n^2 P(X=n) =Var (X) + [E(X)]^2](https://tex.z-dn.net/?f=%5Csum%20%5Climits%20%5E%7Bn%7D_%7Bi%3D1%7D%20n%5E2%20P%28X%3Dn%29%20%3DVar%20%28X%29%20%2B%20%5BE%28X%29%5D%5E2)
∵ X geom(p = 1/2)
Given that during the game play, You pay me $10 , the calculated expected loss = $10 - $6
= $4
∴
In the long run, you expect to lose $4 per game
Answer:
a. 0.0000009 m
b. 6,130, 000, 000
c. 100,000 light years
d. 0.00001 mm
e. 0.0000010 kg
Step-by-step explanation:
Here, we want to come from the standard form to the decimal form
In doing this, we consider the power of 10 attached
If negative, we count leftwards, if positive, we count right-wards
The number of times we count is the power of 10 attached. The count means we are moving the decimal point
For numbers with out visible decimal point, it is at the back of the most right-ward number
Thus, we have these as;
a. 9 * 10^-7
= 0.0000009 m
b. 6.130 * 10^9
= 6,130, 000, 000
c. 1 * 10^5
= 100,000 light years
d. 1 * 10^-5 mm
= 0.00001 mm
e. 10^-7
= 0.0000010 kg
The biggest common factor number is the GCF number. So the greatest common factor 55 and 65 is 5.
