Answer:
7x^2a
Step-by-step explanation:
7x^3a+7x^2a^2
7x^3a = 7 xxxa
7x^2a^2= 7 xxaa
The common terms are 7xxa
7x^2a
This is the greatest common factor
Answer: a) -0.2252, b) 0.8219
Step-by-step explanation:
Since we have given that
Sample size n = 100
Probability that candies are blue = p= 0.26
Probability that company claims that it is blue candy = P = 0.27
So, Q = 1-P= 1-0.27 = 0.73
So, Null hypothesis : 
Alternate hypothesis : 
So, the test statistic would be
Since α = 0.05
So, critical value of z = 1.96
p-value = P(Z>Z(calculated)
Using the excel function , we get that
Hence, a) -0.2252, b) 0.8219
Answer: 15 kilometers per hour
Step-by-step explanation:
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!!!
