<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.
0.00067=67X10^5 so
67X10^5-2.3X10^5
64.7X10^5 which can be expressed as:
6.47X10^6
(not sure if it matters to you or not, but technically we only had two significant figures so it should be rounded to 6.5X10^6)
Y - 3 = -2(x+5)
y - 3 = -2x - 10
+3. +3
y = -2x - 7
Answer:
3 option
Step-by-step explanation:
