<h3>
Answer: 7/10</h3>
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Explanation:
There are 30 days in April. Since it rained 9 of those days, the empirical probability of it raining in April is 9/30 = (3*3)/(3*10) = 3/10.
If we assume that the same conditions (ie weather patterns) hold for May, then the empirical probability of it raining in May is also 3/10. By "raining in May", I mean specifically raining on a certain day of that month.
The empirical probability of it not raining on the first of May is therefore...
1 - (probability it rains)
1 - (3/10)
(10/10) - (3/10)
(10-3)/10
7/10
We can think of it like if we had a 10 day period, and 3 of those days it rains while the remaining 7 it does not rain.
Answer:
y-1=2(x-2)
Step-by-step explanation:
-z³ + 5k^6 + z³ -10k^6
(-z³ cancels out with z³)
5k^6 -10k^6
(then subtract)
Answer is -5k^6
Im gonna guess 6.00 beacause it says round to hundreths
There is no solution for the first one
if you eliminate y you get 2 equations
-13x - 13z = -25
-13x - 13x = -15
- there is no solution to theses
a23 means the element in the second row and the 3rd column
so its -5
