Solution
The 1st term = -3
15th term = 53
15th term = a + d(n-1)
53 = -3 + d(15- 1) [Here n = 15)
53 + 3 = d(14)
14d = 56
d = 4
The common difference is 4.
Answer:
P = 0.008908
Step-by-step explanation:
The complete question is:
The table below describes the smoking habits of a group of asthma sufferers
Nonsmokers Light Smoker Heavy smoker Total
Men 303 35 37 375
Women 413 31 45 489
Total 716 66 82 864
If two different people are randomly selected from the 864 subjects, find the probability that they are both heavy smokers.
The number of ways in which we can select x subjects from a group of n subject is given by the combination and it is calculated as:

Now, there are 82C2 ways to select subjects that are both heavy smokers. Because we are going to select 2 subjects from a group of 82 heavy smokers. So, it is calculated as:

At the same way, there are 864C2 ways to select 2 different people from the 864 subjects. It is equal to:

Then, the probability P that two different people from the 864 subjects are both heavy smokers is:

Answer:
0.91517
Step-by-step explanation:
Given that SAT scores (out of 1600) are distributed normally with a mean of 1100 and a standard deviation of 200. Suppose a school council awards a certificate of excellence to all students who score at least 1350 on the SAT, and suppose we pick one of the recognized students at random.
Let A - the event passing in SAT with atleast 1500
B - getting award i.e getting atleast 1350
Required probability = P(B/A)
= P(X>1500)/P(X>1350)
X is N (1100, 200)
Corresponding Z score = 

I’m pretty sure it’s 3 cause 10 to the power of 3 is 1000
Answer:
12x - 2
Step-by-step explanation:
Step 1: Write expression
4(3x + 2) - 6
Step 2: Distribute 4
12x + 8 - 6
Step 3: Combine like terms
12x - 2