Step-by-step explanation:
so we're making two draws *with* replacement (this is important)
step 1: for the first draw, it wants the probability of getting a sour candy. to calculate this:
(# of sour candy) / (total # of candy)
step 2: for the second draw, it wants the probability of *not* getting a sour candy. to calculate this, you can calculate 1 - (the probability form part 1).
step 3: to find the probability of both events happening together, simply multiply the probabilities from part 1 and 2 together
side note: for step 2, you can only do this because the candy is being replaced. if there were no replacement, you'd have to re-calculate (# of non-sour candies) / (total after the first candy is drawn)
Answer:
C
Step-by-step explanation:
Since this is an indererminate form, use L'Hopital
d(sint)/dt = cos(t)
d[ln(2e^t) - 1] = (2e^t)/[2e^t - 1]
As t --> 0,
cos(0) = 1
(2e^t)/[2e^t - 1] = 2
1/2 is the limit
Answer:
scalene
Step-by-step explanation:
also an acute triangle is something to do with angles so as long as no angles are provided, scalene is the best option
Answer:
4
Step-by-step explanation:
10 * 40%
10 * 0.4 = 4
6a. 1 - 2sin(x)² - 2cos(x)² = 1 - 2(sin(x)² +cos(x)²) = 1 - 2·1 = -1
6c. tan(x) + sin(x)/cos(x) = tan(x) + tan(x) = 2tan(x)
6e. 3sin(x) + tan(x)cos(x) = 3sin(x) + (sin(x)/cos(x))cos(x) = 3sin(x) +sin(x) = 4sin(x)
6g. 1 - cos(x)²tan(x)² = 1 - cos(x)²·(sin(x)²)/cos(x)²) = 1 -sin(x)² = cos(x)²
