I forgot how to convert the inequalities so if someone could please help me that would be great!
1 answer:
WHEN GRAPHING
Step 1. Convert to y = mx + b
- Equation 1: y = 3x/2 - 16/2
- Equation 2: y = -5 - 2x
Step 2. Graph like normal
- Find zeros & plot
- Find y-intercept & plot
Step 3. Shade as indicated by the inequality symbol
- > or ≥ = above line
- < or ≤ = below line
Step 4. If ≤ or ≥ ONLY, then also shade the line
FOR THIS PROBLEM
1. Graph each equation
2. Shade ABOVE line for each
3. Shade line first equation as well
Hope this helps and God bless!
You might be interested in
Answer:
10.93% probability of observing 51 or fewer vapers in a random sample of 300
Step-by-step explanation:
I am going to use the normal approximation to the binomial to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
The standard deviation of the binomial distribution is:
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that ,
.
In this problem, we have that:
So
What is the approximate probability of observing 51 or fewer vapers in a random sample of 300?
Using continuity corrections, this is , which is the pvalue of Z when X = 51.5 So
has a pvalue of 0.1093.
10.93% probability of observing 51 or fewer vapers in a random sample of 300