1.
Chance of finding a bug = 0.35
Chance of not finding a bug = 1 - 0.35 = 0.65
Probability of finding a bug in the first 3 programs =
Probability of not finding a bug in 2 out of the 3 and finding a bug in 1.:
0.65^2 * 0.35 = 0.147 = 0.15
Answer is A.
2.
Probability of heads = 0.50
Probability of tails = 0.50
Probability of heads on the fourth attempt = tails x tails x tails x heads = 0.5 x 0.5 x 0.5 x 0.5 = 0.0625
The answer is B.
Step-by-step answer:
It is not clear what the question exactly wants, but here's the best I can do with possible interpretations.
Let P(t) = population t years after 2000, then
P(t) = (N0)e^(Rt)
where
N0=population of year 2000 = 19169000
R=rate of growth, in ratio = 0.006
e=base of natural logarithm = 2.7182818284...
Then
P(25) = 19169000*e^(0.006*25)
= 19169000*e^(0.15)
= 22271200.59885846
= 22271201
Not sure what it means "the other value", as we only have one.
<span>11.5 Not sure but it should be the answer!!
</span>
Answer:
SAS
Step-by-step explanation:
We can use the SAS postulate because we are given 2 sides with the included angle. If two sides with the included angle of one triangle are congruent to two sides and the included angle of another triangle, the two triangles are congruent.
You can drive 100 miles.
.75m+45=120
.75m=75
M=100
