One example is when n is 11. 11^2 + 4 = 125. 11 is odd but the 11th term of the sequence, 125, is not prime because 125 is divisible by 5 and 25.
Using the normal distribution, we have that:
- The distribution of X is
.
- The distribution of
is
.
- 0.0597 = 5.97% probability that a single movie production cost is between 55 and 58 million dollars.
- 0.2233 = 22.33% probability that the average production cost of 17 movies is between 55 and 58 million dollars. Since the sample size is less than 30, assumption of normality is necessary.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
is given by:

- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation
.
In this problem, the parameters are given as follows:

Hence:
- The distribution of X is
.
- The distribution of
is
.
The probabilities are the <u>p-value of Z when X = 58 subtracted by the p-value of Z when X = 55</u>, hence, for a single movie:
X = 58:


Z = 0.05.
Z = 0.05 has a p-value of 0.5199.
X = 55:


Z = -0.1.
Z = -0.1 has a p-value of 0.4602.
0.5199 - 0.4602 = 0.0597 = 5.97% probability that a single movie production cost is between 55 and 58 million dollars.
For the sample of 17 movies, we have that:
X = 58:


Z = 0.19.
Z = 0.19 has a p-value of 0.5753.
X = 55:


Z = -0.38.
Z = -0.38 has a p-value of 0.3520.
0.5753 - 0.3520 = 0.2233 = 22.33% probability that the average production cost of 17 movies is between 55 and 58 million dollars. Since the sample size is less than 30, assumption of normality is necessary.
More can be learned about the normal distribution at brainly.com/question/4079902
#SPJ1
So here are the answers to the given questions above:
1. Since the equation is already given, what we are going to do is to solve for x first.
<span>x + (2x – 10) + 1/2 (2x – 10) = 145
x + (2x -10) + x - 5 = 145
x + 2x + x - 10 - 5 = 145
4x = 145 +10 + 5
4x = 160 <<divide both sides by 4 and we get
x = 40.
Therefore, the amount of white paint is 40ml.
The amount of blue paint is 70ml and the green paint is 35ml.
So the difference between the a</span><span>mounts of white paint and blue paint Vivian combined is 30ml.
2. T</span>he equation of the line that passes through the points (-2, 1) and (1, 10) would be: <span>3x - y = -7, the last option. In order to check, just plug in the given ordered pairs.</span>