Answer:
y = 80+ 15*x
members = initial membership + growth* months
y = 80+ 15*x
Answer:
D. 274
Step-by-step explanation:
A normal distribution of the scores is assumed. In the figure attached, the standard normal distribution table is shown.
If only top 5% of athletes are part of the team, then we need to find the value of the table which has a probability of 95%, that value is between 1.64 and 1.65, so we interpolate it as 1.645. The table was made for a variable with mean = 0 and standard deviation (sd) equal to 1, therefore to refer the result to our variable we compute:
1.645 = (x - mean)/sd
x = 1.645*sd + mean
x = 1.645*15 + 250 ≈ 274
So, 95% of the scores are below 274, then 274 is the minimum qualifying score
Answer:
12
Step-by-step explanation:
30% is 0.3, 40 x 0.3 . is 12 :DD
Answer:
Step-by-step explanation:
If the standard deviation of an exam is 5, the z-score us 1.95 and the mean is 80, the actual test score is; 89.75
<h3>How to solve z-score problems?</h3>
We are given;
Standard deviation; s = 5
z-score = 1.95
Mean = 80
Formula for z-score is;
z = (x' - μ)/σ
Thus;
1.95 = (x' - 80)/5
1.95 * 5 = (x' - 80)
9.75 = x' - 80
x' = 80 + 9.75
x' = 89.75
Read more about Z-score Problems at; brainly.com/question/25638875
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