Answer: a.) $50188 to $57812
Step-by-step explanation: <u>Confidence</u> <u>Interval</u> (CI) is an interval of values in which we are confident the true mean is in.
The interval is calculated as
x ± 
a. For a 95% CI, z-value is 1.96.
Solving:
54,000 ± 
54,000 ± 
54,000 ± 1.96*1732.102
54,000 ± 3395
This means the interval is
50605 < μ < 57395
<u>With a 95% confidence interval, the mean starting salary of college graduates is between 50605 and 57395 or </u><u>from 50188 to 57812$.</u>
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b. The mean starting salary for college students in 2017 is $50,516, which is in the confidence interval. Therefore, since we 95% sure the real mean is between 50188 and 57812, there was no significant change since 2017.
-2(x+5) OR 2(-x-5) I hope that helps if not please let me know!
Answer:
Step-by-step explanation:
okay so you would go down to -8, then the slope would be 4/1 which means to go up four and then over one. then you have the line.
i hope this helps you!
Taking the derivative of 7 times secant of x^3:
We take out 7 as a constant focus on secant (x^3)
To take the derivative, we use the chain rule, taking the derivative of the inside, bringing it out, and then the derivative of the original function. For example:
The derivative of x^3 is 3x^2, and the derivative of secant is tan(x) and sec(x).
Knowing this: secant (x^3) becomes tan(x^3) * sec(x^3) * 3x^2. We transform tan(x^3) into sin(x^3)/cos(x^3) since tan(x) = sin(x)/cos(x). Then secant(x^3) becomes 1/cos(x^3) since the secant is the reciprocal of the cosine.
We then multiply everything together to simplify:
sin(x^3) * 3x^2/ cos(x^3) * cos(x^3) becomes
3x^2 * sin(x^3)/(cos(x^3))^2
and multiplying the constant 7 from the beginning:
7 * 3x^2 = 21x^2, so...
our derivative is 21x^2 * sin(x^3)/(cos(x^3))^2
