Answer:
Step-by-step explanation:
log8 62 = x1
8^x1= 62
8^2 =64 >62 so <u>x1<2</u>
log7 50=x2
7^x2=50
7^2=49<50 so <u>x2>2</u>
we have, x1<2 and 2<x2
x1<2<x2
x1<x2
log8 62<log7 50
Answer:
B
Step-by-step explanation:
Answer:
Step-by-step explanation:
Given that A be the event that a randomly selected voter has a favorable view of a certain party’s senatorial candidate, and let B be the corresponding event for that party’s gubernatorial candidate.
Suppose that
P(A′) = .44, P(B′) = .57, and P(A ⋃ B) = .68
From the above we can find out
P(A) =
P(B) =
P(AUB) = 0.68 =
a) the probability that a randomly selected voter has a favorable view of both candidates=P(AB) = 0.30
b) the probability that a randomly selected voter has a favorable view of exactly one of these candidates
= P(A)-P(AB)+P(B)-P(AB)
c) the probability that a randomly selected voter has an unfavorable view of at least one of these candidates
=P(A'UB') = P(AB)'
=
Answer:
if repetition is allowed, if repetition is not allowed.
Step-by-step explanation:
For the first case, we have a choice of 26 letters <em>each step of the way. </em>For each of the 26 letters we can pick for the first slot, we can pick 26 for the second, and for each of <em>those</em> 26, we can pick between 26 again for our third slot, and well, you get the idea. Each step, we're multiplying the number of possible passwords by 26, so for a four-letter password, that comes out to 26 × 26 × 26 × 26 = possible passwords.
If repetition is <em>not </em>allowed, we're slowly going to deplete our supply of letters. We still get 26 to choose from for the first letter, but once we've picked it, we only have 25 for the second. Once we pick the second, we only have 24 for the third, and so on for the fourth. This gives us instead a pretty generous choice of 26 × 25 × 24 × 23 passwords.