Answer:
0.0035289
Step-by-step explanation:
From the question;
mean annual salary = $63,500
n = sample size = 31
Standard deviation = $6,200
Firstly, we calculate the z-score of $60,500
Mathematically;
z-score = x-mean/SD/√n = (60500-63500)/6200/√(31) = -2.6941
So we want to find the probability that P(z < -2.6941)
We can get this from the standard normal table
P( z < -2.6941) = 0.0035289
Answer:
Step-by-step explanation:
Simplify 3 ratical- 1000
Answer:
2.82843
Step-by-step explanation:
it was what i got
Answer:
<em>Answer is </em><em>given below with explanations</em><em>.</em><em>,</em>
Step-by-step explanation:

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