<span> 9.187 to the nearest tenth is 9.2
The 1 in 9.187 is in the tenths place and since the number after it is over 5, you round up. </span>
I think it’s the first one
The common factors of 32 and 48 are 16, 8, 4, 2, 1,
The only way 3 digits can have product 24 is
1 x 3 x 8 = 241 x 4 x 6 = 242 x 2 x 6 = 242 x 3 x 4 = 24
So the digits comprises of 1,3,8 or 1,4,6, or 2,2,6, or 2,3,4
To be divisible by 3 the sum of the digits must be divisible by 3.
1+ 3+ 8=12, 1+ 4+ 6= 11, 2 +2 + 6=10, 2 +3 + 4=9Of those sums of digits, only 12 and 9 are divisible by 3.
So we have ruled out all but integers whose digits consist of1,3,8, and 2,3,4.
Meanwhile they must be odd they either must end in 1 or 3.
The only ones which can end in 1 are 381 and 831.
The others must end in 3.
They must be greater than 152 which is 225. So the
First digit cannot be 1. So the only way its digits can contain of1,3,8 and close in 3 is to be 813.
The rest must contain of the digits 2,3,4, and the only way they can end in 3 is to be 243 or 423.
So there are precisely five such three-digit integers: 381, 831, 813, 243, and 423.
For a binomial experiment in which success is defined to be a particular quality or attribute that interests us, with n=36 and p as 0.23, we can approximate p hat by a normal distribution.
Since n=36 , p=0.23 , thus q= 1-p = 1-0.23=0.77
therefore,
n*p= 36*0.23 =8.28>5
n*q = 36*0.77=27.22>5
and therefore, p hat can be approximated by a normal random variable, because n*p>5 and n*q>5.
The question is incomplete, a possible complete question is:
Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us.
Suppose n = 36 and p = 0.23. Can we approximate p hat by a normal distribution? Why? (Use 2 decimal places.)
n*p = ?
n*q = ?
Learn to know more about binomial experiments at
brainly.com/question/1580153
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