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.
Y=a(x-h)^2+k
is what vertex form look like. (h,k) is your vertex so in this problem your vertex would be (60,200)
Answer:
Option E is correct.
The expected number of meals expected to be served on Wednesday in week 5 = 74.2
Step-by-step Explanation:
We will use the data from the four weeks to obtain the fraction of total days that number of meals served at lunch on a Wednesday take and then subsequently check the expected number of meals served at lunch the next Wednesday.
Week
Day 1 2 3 4 | Total
Sunday 40 35 39 43 | 157
Monday 54 55 51 59 | 219
Tuesday 61 60 65 64 | 250
Wednesday 72 77 78 69 | 296
Thursday 89 80 81 79 | 329
Friday 91 90 99 95 | 375
Saturday 80 82 81 83 | 326
Total number of meals served at lunch over the 4 weeks = (157+219+250+296+329+375+326) = 1952
Total number of meals served at lunch on Wednesdays over the 4 weeks = 296
Fraction of total number of meals served at lunch over four weeks that were served on Wednesdays = (296/1952) = 0.1516393443
Total number of meals expected to be served in week 5 = 490
Number of meals expected to be served on Wednesday in week 5 = 0.1516393443 × 490 = 74.3
Checking the options,
74.3 ≈ 74.2
Hence, the expected number of meals expected to be served on Wednesday in week 5 = 74.2
Hope this Helps!!!
9 packets.
36 divided by 3 = 12
49 divided by 5 = 9.8
76 divided by 7 = 10.8
Maximum packets that can be obtained is 9 packets because only 9 packets with 5 green beads each can be packaged.